(*) Introduction:
Consider a Kho-Kho court of dimension Legth(29m) * 16m(Breadth).This is to be a rectangular enclosure of area 558 m^2.
They want to leave space of equal width all around the court for the spectators.
What would be the width of the space for spectators? would it be enough?
Suppose the width of the space be 'x' meter. So from the figure
length of the plot would b (29+2x) meter.
Breadth of the rectangular plot would be (16+2x) m
.^., area of the rectangular plot
will be = length * breadth
= (29+2x) * (16 +2x)
Since the area of the plot is = 558 m^2
.^. (29+2x)(16+2x) = 558
.^. 4x^2 + 90x + 464 = 558
=> 4x^2 + 90x - 94 = 0 (dividing by 2)
=> 2x^2 + 45x - 47 = 0 ---(1)
The value of 'x' from the above equation will give the possible width of the space for spectators.
Let us consider another example :
Rani has a square metal sheet. She removed squares of side 9 cm, from each corner of this sheet. Of the remaining sheet, she turned up the sides to form an open box. The capacity of the box is 144 cc. Can we find out the dimensions of the metal sheet?
Suppose the side of the square piece of metal sheet be 'x' cm.
Then, the dimensions of the box are
9 cm. * (x-18) cm. * (x-18) cm.
Since volume of the box is 144 cc
9(x-18)(x-18) = 144
(x-18)^2 = 16
x^2 - 36x + 308 = 0
So, the side 'x' of the metal sheet will satisfy the equation.
x^2 - 36x + 308 ---------(2)
Let us observe the L.H.S of equation (1) and (2)
Are they quadratic polynomials?
Since, the LHS of the above equations are quadratic polynomials they are called quadratic equations.
(*) Quadratic Equations
A quadratic equation in the variable 'x' is an equation of the form ax^2 +bx =c = 0, where a,b,c are real numbers and a=/=0.
Example: 2x^2 + x - 300 = 0 is quadratic equation.
Similarly,
2x^2 -3x +1 =0
4x-3x^2+2 = 0 and
1 - x^2 + 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x)=0, where p(x) is polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation.
That is, ax^2+bx+c=0, a=/=0 is called standard form of a quadratic equation and
y = ax^2+bx+c is called a quadratic equation.
There are various uses o Quadratic functions. Some of them are :-
1) When the rocket is fired upward, then the height of the rocket is defined by a ' quadratic equation.'
2) Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses and orbits of the celestial objects are defined by the quadratic equations.
3) The path of a projectile is defined by quadratic function.
4) When the breaks are applied to a vehicle, the stopping distance is calculated by using quadratic equation.
Example-1. Represent the following situations mathematically:
(!) Raju and Rajendar together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles now they have is 124. We would like to find out how many marbles they had previously.
sol) Let the number of marbles Raju had be 'x'
Then the number of marbles
Rajendar had = 45-x
The number of marbles left with Raju, when he lost 5 marbles = x- 5
The number of marbles left with rajendar, when he lost 5 marbles= (45-x)-5
=> 40-x
Therefore, their product = (x-5)(40-x)
=> 40x - x^2 - 200 +5x
=> -x^2 +45x -200 =124(Given product =124)
=> x^2 -45x +324 = 0 (multiply -ve sign)
Therefore, the number of marbles raju had 'x', satisfies the quadratic equation
x^2 - 45x +324 = 0
Which is the required representation of the problem mathematically.
(!!) The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the other two sides is 5 cm. We would like to find out the length of the two sides?
sol)
Let the length of smaller side be 'x' cm.
Then length of larger side = (x+5) cm.
Given length of hypotenuse = 25 cm.
In a right angle triangle we know that
(hypotenuse)^2 = (side)^2 + (side)2
So,
(25)^2 = x^2 + (x+5)^2
625 = x^2 + x^2 +25 +10x
2x^2 + 10x - 600 = 0
x^2 +5x - 300 = 0
Value of 'x' from the above equation will give the possible value of length of sides of the given right angled triangle.
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Example-2. Check whether the following are quadratic equations.
(!) (x-2)^2 + 1 = 2x - 3
sol) LHS = (x-2)^2 + 1
=> x^2 -4x +4 +1
=> x^2 -4x +5
.^., (x-2)^2 +1 = 2x-3 can be written as
x^2 - 4x + 5 = 2x - 3
i.e, x^2 - 6x + 8 = 0
It is in the form of = ax^2 + bx + c = 0
Therefore, the given equation is a quadratic equation.
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(!!) x(x+1) +8 = (x+2)(x-2)
sol) LHS = x(x+1) + 8 = x^2 +x+8
RHS = (x+2)(x-2) = x^2 - 4
Therefore,
x^2 + x + 8 = x^2 - 4
x^2 +x+8-x^2 +4 = 0
i.e, x + 12= 0
It is not in the form = ax^2 +bx+c = 0
Therefore, the given equation is not a quadratic equation.
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(!!!) x(2x+3) = x^2 + 1
sol) LHS = 2x^2 + 3x
So, x(2x+3) = x^2 +1 can be written as
2x^2 + 3x = x^2 + 1
Therefore, we get
x^2 + 3x - 1 = 0
It is in the form of= ax^2 +bx+c = 0
So, the given equation is a quadratic equation.
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(!V) (x+2)^3 = x^3 - 4
sol) LHS = (x+2)^2 * (x+2)
=> (x^2 + 4x + 4 )* (x+2)
=> x^3 + 2x^2 + 4x^2 + 8x + 4x + 8
=> x^3 + 6x^2 + 12x + 8
Therefore,(x+2)^3 = x^3 -4 can be written as
x^3 + 6x^2 + 12x + 8 = x^3 - 4
i.e, 6x^2 + 12x + 12 = 0
or
x^2 + 2x + 2 = 0
It is in the form of = ax^2 + bx + c = 0
So, the given equation is a quadratic equation,
Remark : In (!!) above, the given equation appears to be a quadratic equation, but is not a quadratic equation.
In (!V) above, the given equation appears to be a cubic equation( an equation of degree 3 ) and not a quadratic equation.
But it turns out to be a quadratic equation.As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.
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Exercise - 5.1
1) check whether the following are quadratic equations
(1) (x +1) ^2 = 2 (x -3)
sol)
(x +1) ^2 => (a+b)^2
x^2 + 1^2 + 2 *x*1 => a^2 + b^2 + 2ab
Now,
x^2 + 1 + 2x = 2(x - 3)
x^2 + 1 + 2x = 2x - 6
x^2 + 1 + 2x -2x +6 = 0
x^2 + 7 = 0
We can write,
x^2 + 0x + 7 = 0 { a^2 + bx + c form]
where a = 1 , b = 0, c = 7
Hence, it is quadratic equation.
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2) x^2 - 2x = (-2) (3 -x)
sol) x^2 - 2x = (-2 * 3) - (-2* x)
=> x^2 - 2x = -6 +2x
=> x^2 - 2x +6 - 2x = 0
=> x^2 -4x + 6 = 0 { ax^2 + bx + c form}
where a = 1, b = -4, c = 6
Hence, it is quadratic equation.
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3) (x - 2) (x +1) = (x - 1) (x +3)
sol) x* ( x +1) -2 *(x + 1) = x ( x+3) -1 ( x +3)
=>x^2 + x - 2x - 2 = x^2 + 3x - x - 3
=> x - 2x - 2 = 3x - x - 3
=> -x - 2 = 2x -3
=> -x - 2x -2 + 3
=> -3x + 1 = 0 {not equal to =/= ax^2 + bx + c }
Hence, is it not a quadratic equation.
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4) (x - 3) ( 2x +1) = x (x+5)
sol) x ( 2x + 1) - 3 (2x +1) = x (x+5)
=>2x^2 + x - 6x - 3 = x^2 + 5x
=> x^2 + x - 6x -3 = 5x
=> x^2 - 5x - 3 = 5x
=> x^2 - 5x - 5x - 3 = 0
=> x^2 - 10x - 3 = 0 { ax^2 + bx + c = 0 form}
where a = 1, b = -10, c = -3
Hence its a quadratic equation.
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5) (2x - 1 ) ( x - 3) = (x +5 ) (x - 1)
sol) 2x ( x - 3) -1 (x - 3) = x ( x - 1) + 5 (x - 1)
=>2x^2 - 6x - x + 3 = x^2 - x + 5x - 5
=> x^2 - 6x -x + 3 = -x + 5x - 5
=> x^2 - 7x + 3 = 4x - 5
=>x^2 -7x - 4x +3 + 5 = 0
=>x^2 - 11x + 8 = 0 { ax^2 + bx + c form]
where a = 1, b = -11, c = 8
Hence its a quadratic equation.
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6) x^2 + 3x + 1 = (x - 2) ^2
sol) x^2 + 3x + 1 = x^2 + 2^2 - 2 * x * 2
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compare (x-2)^2 with ( a-b)^2 = a^2 + b^2 - 2ab
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=> x^2 + 3x + 1 = x^2 + 4 - 4x
=> 3x + 1 = 4 - 4x
=> 3x + 1 - 4 + 4x = 0
=> 3x - 3 + 4x = 0
=> 7x - 3 = 0 { =/= ax^2 + bx + c}
Hence, it is not a quadratic solution.
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7) ( x + 2) ^3 = 2x ( x^2 - 1)
sol) x ^3 + 2^3 + 3*x*2 (x+2) = 2x ( x^2 - 1)
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compare (x +3)^3 with (a +b)^3 = a^3 + b^3 + 3ab(a+b)
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=> x ^3 + 2^3 + 3*x*2 (x+2) = 2x^3 - 2x
=> x^3 + 8 + 6x(x + 2) = 2x^3 - 2x
=> x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x
=> x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 = 0
=> -x^3 + 6x^2 + 14x + 8 = 0 { =/= ax^2 + bx +c}
Hence, it is not a quadratic equation.
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8) x^3 - 4x^2 - x + 1 = ( x - 2)^3
sol) x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3*x*2(x - 2)
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compare (x - 2)^3 with (a-b)^3 = a^3 - b^3 - 3ab(a - b)
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=> x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x(x -2)
=>x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x^2 + 12x
=>-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0
=> 2x^2 - 13x + 9 = 0 { == ax^2 + bx + c}
where a = 2, b = -13 and c = 9
Hence, it is a quadratic equation.
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2) Represent the following situation in the form of quadratic equation.
1) The area of a rectangular plot is 528m.sq . The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
sol) Given that Area = 528m.sq
condition 1:- length of the plot is one more than twice its breadth
let length be = L
Let breadth be = b
So, length (L) = 2b + 1
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We have , Area of rectangle = length * breadth
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=> 528 = (2b + 1) * b
=> 528 = 2b^2 + b
=>2b^2 + b - 528 = 0 { ax^2 + bx + c=0}
where a = 2, b = 1, c = -528
Hence, it is quadratic equation.
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2) The product of two consecutive positive integers is 306. We need to find the integers.
sol) Let 1st number be = x
Let 2nd number be = x +1
condition :- product of two consecutive integers = 306
=> x * (x+1) = 306
=> x^2 + x = 306
=> x^2 + x - 306 { ax^2 + bx + c form}
where a = 1 , b = 1, c = -360
Hence, it is a quadratic equation.
=> x ( x +29) + 3 ( x + 29) = 360
=> x^2 + 29x + 3x + 87 = 360
=> x^2 + 29x + 3x + 87 - 360 = 0
=> x^2 + 32x - 273 = 0 { == ax^2 + bx + c form }
where a = 1 , b = 32, c = -273
Hence, it is a quadratic equation.
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4) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hrs more to cover the same distance. We need to find the speed of the train.
sol) Let the speed of train be = x km/hr
Condition 1 :- Train travels 480 km distance at uniform speed.
Distance = 480
Speed = x km/h
We have
Speed = Distance
Time
x = 480
Time
Time = 480
x --------- (1)
Condition 2 :- Speed had been 8km/h less
Distance = 480 km
Speed = ( x - 8) km/hr
3 hr more to cover :- 480 + 3
x
We have
Speed = Distance
Time
(x - 8) = 480
( 480 + 3 )
x
(x- 8) ( 480 + 3 ) = 480 --------------- (2)
x
Solving (2) we get
=> (x - 8) ( 480 + 3 ) = 480
x
=> (x - 8) ( 480 + 3x ) = 480
x
=> (x - 8) ( 480 + 3x) = 480x
=> x ( 480 + 3x) - 8 ( 480 + 3x) = 480x
=>480x + 3x^2 - 3840 - 24x - 480x = 0
=> 3x^2 - 24x - 3840 = 0
=> 3 ( x^2 - 8x - 1280) = 0
=> x^2 - 8x - 1280 = 0 { ==ax^2 + bx + c form}
Where a = 1, b= -8, c = -1280
Hence , its quadratic equation
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(*) Solutions of a Quadratic Equation By factorization:
Consider the quadratic equation 2x^2-3x+1=0. If we replace 'x' by '1'. Then, we get
(2*1^2) -(3*1)+1 = 0 = RHS of the equation.
Since 1 satisfies the equation, we say that 1 is a root of the quadratic equation
2x^2 - 3x + 1= 0.
.^. x=1 is a solution of the quadratic equation.
This also means that 1 is a zero of the quadratic polynomial 2x^2-3x+1.
In general, a real number 𝜶 is called a root of the quadratic equation ax^2 + bx + c = 0.
If a𝜶^2 + b𝜶 + c = 0. We say that x =𝜶 is a solution of the quadratic equation, or 𝜶 satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax^2+bx+c and the roots of the quadratic equation ax^2+bx+c=0 are the same.
We observed that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have atmost two roots.
Example-3. Find the roots of the equation 2x^2-5x+3=0, by factorization.
sol) Let us first split the middle term. recall that if ax^2+bx+c is a quadratic equation polynomial then to split the middle term we have to find two numbers p and q such that
p+q=b and p*q= a*c
So to split the middle term of 2x^2 -5x +3, we have to find two numbers p and q such that
p+q= -5 and p*q = 2*3=6
For this we have to list out all possible pair of factors of 6.
They are (1,6);(-1,-6);(2,3);(-2,-3) . from the list it is clear that the pair (-2,-3) will satisfy our condition p+q=-5 and p*q=6.
The middle term '-5x' can be written as '-2x-3x'.
So, 2x^2 -5x + 3 = 2x^2-2x-3x+3
= 2x(x-1) -3 (x-1) = (2x-3)(x-1)
Now, 2x^2 -5x +3 = 0 can be rewritten as
(2x-3)(x-1) = 0.
So, the values of x for which 2x^2-5x+3=0 are the same for which (2x-3)(x-1) = 0.
i.e, either 2x-3 =0 or x -1=0
Now,
2x-3=0 gives x = 3/2
and
x-1 =0 gives x =1
So, x=3/2 and x=1 are the solutions of the equation.
In other words, 1 and 3/2 are the roots of the equation 2x^2-5x+3=0
Note: We have found the roots of 2x^2-5x+3=0 by factorizing 2x^2-5x+3 into two linear factors and equating each factor to zero.
Example-4. Find the roots of the quadratic equation x - 1/3x = 1/6
sol) 3x^2 - 1 = 1
3x 6
=> 18x^2 - 6 = 3x
=> 18x^2 -3x - 6 = 0 (divide by 3)
=> 6x^2 - x -2 = 0
=> 6x^2 +3x-4x -2 = 0
=> 3x(2x +1) -2(2x+1)
=> (3x-2)(2x+1)
The roots of 6x^2-x-2=0 are the values of x for which (3x-2)(2x+1) = 0
Therefore, 3x-2=0 and 2x+1=0
i.e, x = 2/3 or x = -1/2
Therefore, the roots of 6x^2-x-2 =0 are 2/3 and -1/2
We verify the roots, by checking that 2/3 and -1/2 satisfy 6x^2-x-2=0.
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Example-5. Find the width of the space for spectators in kho-kho ground as we discussed above
sol) We found that if the width of the space for spectators is x m., then x satisfies the equation 2x^2 +45x-47=0. Applying the factorization method we write this equation as :-
2x^2 - 2x + 47x - 47 = 0
2x(x-1) + 47(x-1) = 0
i.e, (x-1)(2x+47) = 0
So, the roots of the given equation are x = 1 or x = -47/2. Since 'x' is the width of space of the spectators it cannot be negative.
Thus, the width is 1 m.
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Exercise- 5.2
1) Find the roots of the following quadratic equations by factorisation ;
1) x ^2 = 3x - 10 = 0
sol) x^2 - 2x - 5x - 10 = 0
=> x ( x +2) - 5 ( x + 2) = 0
=> (x - 5) (x +2) = 0
=> x = 5 or x = -2
Hence, x = 5, -2 are the roots of equation.
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2) 2x^2 + x - 6 = 0
sol) 2x^2 + 4x -3x - 6 = 0
=> 2x ( x + 2) - 3 ( x + 2) = 0
=> (2x - 3) ( x +2) = 0
=> 2x - 3 = 0 (x+2) = 0
=> 2x = 3 x = -2
=> x = 3 x = -2
2
Hence x = 3 / 2 , - 2 are the roots of equation.
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3) _/2 x^2 + 7x + 5_/2 = 0
sol) _/2x^2 + 2x + 5x + 5_/2 = 0
=> _/2x^2 + (_/2 * _/2)x + 5x + 5_/2 = 0
=> _/2x(x + _/2) + 5(x + -/2) = 0
=> (_/2x + 5) (x + -/2) = 0
=> _/2x + 5 = 0 x + -/2 = 0
=> _/2x = -5 x = - _/2
=> x = -5
_/2
Hence, x = -5 , - _/2 are the roots of equation.
_/2
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4) 2x^2 - x + 1 = 0
8
sol)( 8 * 2x^2) - (8*x ) + 1 = 0
8
=> 16x^2 - 8x + 1 = 0 / 8
=> 16x^2 - 4x - 4x + 1 = 0
=> 4x(4x - 1) -1(4x - 1) = 0
=> (4x - 1) (4x -1) = 0
=> 4x - 1 = 0 4x - 1 = 0
=> 4x = 1 and 4x = 1
=>x = 1 / 4 and x = 1 / 4
Hence, x = 1 /4, 1/4 are the roots of equation.
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5) 100x^2 - 20x + 1 = 0
sol) 100x^2 - 10x - 10x + 1 = 0
=> 10x(10x - 1) -1(10x - 1) = 0
=> (10x-1) (10x - 1) = 0
=> 10x = 1 and 10x = 1
=> x = 1/10 and x = 1 / 10
Hence, x = 1 /10, 1/10 are the roots of equation.
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6) x(x + 4) = 12
sol) x^2 + 4x = 12
=> x^2 + 4x -12 = 0
=> x^2 + 6x - 2x - 12 = 0
=> x(x + 6) -2( x + 6) = 0
=> (x - 2 ) ( x +6) = 0
=> x -2 = 0 and x +6 = 0
=> x = 2 and x = -6
Hence, x = 2, -6 are the roots of equation.
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7) 3x^2 - 5x + 2 = 0
sol) 3x^2 - 3x - 2 x + 2 = 0
=> 3x ( x - 1) -2 ( x - 1) = 0
=> (3x - 2) (x - 1) = 0
=> 3x - 2 = 0 and x - 1 = 0
=> x = 2 / 3 and x = 1
Hence, x = 2 /3, 1 are roots of the euqtion.
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8) x - 3 = 2
x
sol) x*x - 3 = 2
x
=> x^2 - 3 = 2 *x
=> x^2 - 2x - 3 = 0
=> x^2 - 3x + x - 3 = 0
=> x(x - 3) + 1 (x -3) = 0
=> (x - 3)(x + 1) = 0
=> x = 3 and x = -1
Hence, x = 3, -1 are the roots of the equation.
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9) 3 (x - 4)^2 - 5(x - 4) = 12
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sol) compare (x - 4) ^2 with (a - b )^2 = a^2 +b^2 - 2ab
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=> 3 (x^2 + 4^2 - 2*x*4 ) -5x + 20 = 12
=> 3( x^2 + 16 -8x) - 5x + 20 - 12 = 0
=> 3x^2 + 48 - 24x - 5x + 8 = 0
=> 3x^2 - 29x + 56 = 0
=> 3x^2 -21x - 8x + 56 = 0
=> 3x ( x - 7) - 8 ( x -7)
=> (3x - 8)(x - 7) = 0
=> 3x - 8 = 0, and x - 7 = 0
=> x = 8 / 3 and x = 7
.^. x = 8/3, 7 are the roots of the equation.
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2) Find two number whose sum is 27 and product is 182
sol) let the 1st number be = x
let the 2nd number be = y
Given :- sum of two number = 27
=> x + y = 27 ---- (1)
Given :- product of two number = 182
=> x * y = 182 ------ (2)
using eq.1.. we get y = 27 - x
substitute "y" in eq.2 we get
=> x * y = 182
=> x * (27 - x) = 182
=> 27x - x^2 - 182
=> -x^2 + 27x - 182 = 0
=> x^2 - 27x + 182 = 0
=> x^2 - 13x - 14x + 182 = 0
=> x(x - 13) - 14( x - 13) = 0
=> (x - 14) ( x - 13) = 0
=> x = 14 and x = 13
When x = 13
put "x" value in => x + y = 27
=> 13 + y = 27
=> y = 27 - 13
=> y = 14
.^. when 1st number = 13
2nd number = 14
When x = 14
=> x = 27 - 14
=> x = 13
.^. when 1st number = 14
2nd number = 13
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3) Find two consecutive positive integers, sum of whose squares is 613
sol) let 1st number = x
Let 2nd number = x +1
Condition :- sum of squares = 365
.i.e, x^2 + (x + 1) ^2 = 365
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compare (x + 1)^2 with (a + b)^2 = a^2 + b^2 + 2ab
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=> x^2 + ( x^2 + 1^2 + 2*x*1) = 365
=> 2x^2 + 1 + 2x = 365
=> 2x^2 + 2x - 365 + 1 = 0
=> 2x^2 + 2x - 364 = 0
=> 2 (x^2 + x - 182) = 0
=> x^2 + x - 182 = 0 / 2
=> x^2 + 14x - 13x - 182 = 0
=> x ( x + 14) - 13( x + 14) = 0
=> (x - 13) (x + 14) = 0
=> x = 13 and x = -14
Since its given in question it is positive integers, we can take only " x = 13" .We cannot take negative "x = -14"
When(first no) x = 13
Then 2nd number = x + 1 => 13 + 1 => 14
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4) the altitude of a right traingle is 7cm less than its base . if the hypotenuse is 13cm, find the other two sides.
sol) Let base of the triangle = x
condition : - the altitude(height) is 7cm less than base(i.e, "x")
.^. height = x- 7
Since It is a right angled triangle
using Pythagoras theorem
Formula : - ( Hypotenuse )^2 = ( height) ^2 + (Base) ^2
=> (AC)^2 = (AB)^2 + (BC)^2
Now,We have:-
Hypotenuse(AC) = 13cm ----(1)
Height (AB) = x - 7 ----(2)
Base( BC) = X ----(3)
Substitute (1), (2), (3) value in formula we get,
> (AC)^2 = (AB)^2 + (BC)^2
=> (13)^2 = (x - 7)^2 + x^2
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Note:- (x-7)^2 is in (a-b)^2 ={ a^2 + b^2 - 2ab} form
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=> 13 * 13 = (x^2 + 7^2 - 2(x)(7)) + x^2
=> 169 = x^2 + 49 - 14x + x^2
=>x^2 + x^2 - 14x +49 - 169 = 0
=> 2x^2 -14x - 120 = 0
=> 2(x^2 - 7x - 60) = 0
=> x^2 - 7x - 60 = 0
2
=> x^2 + 5x - 12x - 60 = 0
=> x(x + 5) -12(x + 5) = 0
=> (x - 12)(x + 5) = 0
=> x- 12 = 0 and x +5 = 0
=> x = 12 and x = -5
Since, height cannot be in negative unit we take just x = 12
.^. height = x - 7
= 12 - 7
= 5
and Base = 12
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5) A cottage industry produces a certain number of pottery articles in a day. it was observed on a particular day that the cost of production of each article ( in rupees) was 3 more than twice the number of articles produced on that day. if the total cost of production on that day was Rs,90, find the number of articles produced and the cost of each article.
Sol) Given :- total cost of production = 90Rs
Let the number of articles = x
condition:- cost of production of each article( rupees) was 3 more than twice the number of articles produced on that day
i.e, cost of article = twice the number of articles (2x) + three more (3)
= 2x + 3
We know
Total cost of production = number of articles produced * Cost of article
=> 90 = x*(2x+3)
=> 90 = 2x^2 + 3x
=> 2x^2 + 3x - 90 = 0
=> 2x^2 + 15x - 12x - 90 = 0
=> x ( 2x + 15) -6(2x + 15) = 0
=> (x - 6) (2x + 15) = 0
=> x-6 = 0 and 2x + 15 = 0
=> x = 6 and 2x = -15 => x = -15 / 2
But cost cannot be in negative . thus we take x = 6
.^. Number of articles = 6
and cost of articles = 2x + 1
=2(6) + 1
= 13
.^. Cost of articles = 13rs
****************************************************
6) Find the dimensions of a rectangle whose perimeter is 28mts.and whose area is 40.sq.mt
sol) let length of rectangle be = L
Let breadth of rectangle be = b
Given :- Perimeter is 28mts.
Formula :- perimeter of rectangle (28) = 2(L + b)
=> 2(L + b ) = 28
=> L + b = 28
2
=> L + b = 14
=> b = 14 - L -----------(1)
Given :- Area of rectangle = 40.sq.mt
Formula:- Area of rectangle(40) = L * b
=> 40 = L * b
=> 40 = L * (14 - L)
=> 40 = 14L - L^2
=> 40 - 14l + L^2 = 0
=> L^2 - 14L + 40 = 0
=> l^2 - 10L - 4L + 40 = 0
=> L(L - 10) -4( L - 10) = 0
=> ( L - 4) ( L - 10)
=> L - 4 = 0 or L - 10 = 0
=> l = 4 or L = 10
Substitute "L = 10" in "b= 14 - L"
We get,
=> b = 14 - 10
=> b = 4
Hence , length = 10m and breadth = 4m.
**********************************************************
7) The base of a triangle is 4cm longer than its altitude. if the area of the traingle is 48.sq.cm then find its base and altitude.
sol) Let the altitude be = y
given :- Base is 4cm longer than its altitude
i.e, base = 4 + altitude
= 4 + y
We have area of traingle formula :-
Formula :-
Area of traingle( 48) = 1/2 * base * altitude( height)
=> 48 = 1 * ( 4 + y) * y
2
=> 48 * 2 = 4y + y^2
=> y^2 + 4y - 96 = 0
=> y^2 + 12y - 4y - 96 = 0
=> y(y + 12) -8(y + 12) = 0
=> (y - 8) ( y + 12) = 0
=> y - 8 = 0 or y + 12 = 0
=> y = 8 or y = -12
As Altitude(height) cannot have negative value we ignore y = -12
Now, when y = 8
base = 4 + height
= 4 + 8
= 12
Hence, base = 12 and altitude = 8cm
**************************************************************
8) Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km apart find the average speed of each train.
sol) Let the speed of 1st train be = x km/hr
condition :- 1st train travels 5 km/hr faster than 2nd train.
.^. Speed of 2nd train be = (x - 5) km/hr
condition 2:- After 2 hr they are 50km apart
The first train travels from "O" to "A" towards west
The second train travels from O to B.
Given that AB = 50 km
It is clear that triangle OAB is a right triangle.
using Pythagoras Theorem, We have,
AB^2 = OA^2 + OB^2 ------ (1)
We know that distance = speed * time
=> OA = x * 2 = 2x -----(2)
similarly, we have,
=> OB = 2 (x - 5) ---------(3)
Substitute (2) and (3) in (1) we get
Given distance between them (AB) = 50 km
AB^2 = OA^2 + OB^2
(50)^2 = (2x)^2 + [2(x-5)]^2
2500 =4x^2 + 4 *{ (x-5)]^2
2500 = 4x^2 + 4 (x^2 + 5^2 - 2 * x *5 )
2500 = 4x^2 + 4 ( x^2 + 25 - 10x)
2500 = 4x^2 + 4x^2 + 100 - 40x
2500 = 8x^2 - 40x + 100
=> 8x^2 - 40x + 100 - 2500 = 0
=> 8x^2 - 40x - 2400 = 0
=>8 ( x^2 - 5x - 300) = 0
=> x^2 - 5x - 300 = 0
8
=> x^2 - 5x - 300 = 0
=> x^2 - 20x + 15x - 300 = 0
=> x(x - 20) +15(x - 20) = 0
=> (x + 15) (x - 20) = 0
=> x + 15 = 0 or x - 20 = 0
=> x = -15 or x = 20
Speed cannot be negative. ignore x = -15
Thus, the speed of the first train is = 20 km/hr
Speed of 2nd train (x -5) = 20 - 5 = 15 km/hr
****************************************************************
9).. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was 1600rs. How many boys are there in the class?
sol) Let number of girls be = x
Let number of boys be = y
Total number of student( x + y) = 60
x + y = 60 ----- (1)
condition :- each boy(x) contributed rupees equal to the number of girls(y) and each girl (y) contributed rupees equal to the number of boys(x)
Total money collected = 1600
xy + yx = 1600
=> 2xy = 1600
=> xy = 1600 / 2
=> xy = 800 -------(2)
From eq,(1) we get, y = 60 -x
substitute (y = 60 -x) in (eq. 2)
=> xy = 800
=> x * (60 - x) = 800
=> 60x - x^2 = 800
=> x^2 - 60x + 800 = 0
=> x^2 - 20x - 40x + 800 = 0
=> x(x - 20) -40(x - 20) = 0
=> (x - 40)(x - 20) = 0
=> x - 40 = 0 or x - 20 =0
=> x = 40 or x = 20
if the number of girls is 20, then the number of boys is 40.
And if the number of girls is 40, Then the number of boys is 20
**************************************************************
10) A motor boat heads upstream a distance of 24km on a river whose current is running at 3 km per hour. The trip up and back takes 6hrs. Assuming that the motor boat maintained a constant speed, what was its speed?
sol) let the speed of the boat be = x km/hr
Speed of the boat upstream = (x - 3) km/hr
Speed of the boat downstream = ( x + 3) km/hr
We have formula :- Time = Distance
Speed
Time taken to go upstream = Distance
Speed.
= 24
x - 3
similarly time taken to downstream = Distance
Speed
= 24
x + 3
Condition :- It took total 6hrs
24 + 24 = 6
x - 3 x + 3
=> 24 ( x + 3) + 24 ( x - 3) = 6 (x - 3) (x +3)
=> 24x + 72 + 24x - 72 = 6 (x^2 - 3^2)
=> 48x = 6(x^2 - 9)
=> 48x = 6x^2 - 54
=> 6x^2 - 48x - 54 = 0
=> 6 (x^2 - 8x - 9) = 0
=> x^2 - 8x - 9 = 0
6
=> x^2 - 8x - 9 = 0
=> x^2 -9x +x - 9 = 0
=> x(x - 9) + 1 ( x - 9) = 0
=> (x + 1)(x - 9) = 0
=> x+ 1 = 0 or x - 9 = 0
=> x = -1 or x = 9
Since "x" is speed of the stream, it cannot be negative. Ignore x = -1
.^. the speed of the stream is = 9 km/hr
*****************************************************************
(*) Solution of a quadratic equation by completing the square
Is method of factorization applicable to all types of quadratic equation?
let us try to solve x^2+4x-4=0 by factorization method
To solve the given equation x^2 +4x-4=0 by factorization method.
we have to find 'p' and 'q' such that
p+q= 4 and
p*q = -4
But is it not possible. So by factorization method we cannot solve the given equation.
Therefore, we shall study another method.
Consider the following situation
The product of Sunita's age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this,
let her present age(in years) be x years.
Age before two year = x-2 &
age after four years = x +4
then the product of both the ages is (x-2)(x+4)
Therefore, (x-2)(x+4) = 2x + 1
i.e., x^2 + 2x -8 = 2x + 1
i.e., x^2 - 9 = 0
So, Sunita's present age satisfies the quadratic equation x^2-9=0
We can write this as x^2= 9.
Taking square roots, we get x =3 or x=-3.
Since the age is a positive number, x = 3.
So, Sunita's present age is 3 years.
Now consider another equation (x+2)^2 -9=0.
To solve it, we can write it as (x+2)^2 = 9
Taking square roots, we get
x+2 = 3 or x + 2 = -3
Therefore x = 1 or x = -5
So, the roots of the equation
(x+2)^2 -9 =0 are 1 and -5.
In both the examples above, the term containing x is completely a square, and we found the roots easily by taking the square roots.
But, what happens if we are asked to solve the equation x^2+4x-4=0. And it cannot be solved by factorization also.
So, we now introduce the method of completing the square. The idea behind this method is to adjust the left side of the quadratic equation so that it becomes a perfect square.
The process is a s follows.
x^2 + 4x - 4 = 0
=> x^2 + 4x = 4
=> x^2 + 2x .2 = 4
Now, the LHS is in the form of a^2 + 2ab. If we add b^2 it becomes as a^2 + 2ab = b^2 which is perfect square. So, by adding b^2 = 2^2 =4 to both sides we get,
x^2 + 2x. 2 + 2^2 = 4 + 4
=> (x+2)^2 = 8
=> x + 2 = +√8
=> x = -2 + 2√2
Now consider the equation 3x^2 -5x +2 =0.
Note that the coefficient of x^2 is not 1.
So we divide the entire equation by 3 so that the coefficient of x^2 is 1
.^. x^2 - 5 x + 2 = 0
3 3
=> x^2 - 5 x = -2
3 3
=> x^2 - 2.x.5 = -2
6 3
=> x^2 - 2.x.5 + (5 )^2 = -2 + (5)^2
6 (6)^2 3 (6)^2
( add (5/6)^2 both side )
( x - 5 )^2 = -2 + 25
6 3 36
(a-b)^2 form
(x - 5 )^2 = (12 * -2)+(25*1)
6 36
(x - 5)^2 = -24 +25
6 36
(x - 5 )^2 = 1
6 36
x - 5 = + 1
6 6
So, x = 5 + 1 or x = 5 - 1
6 6 6 6
Therefore, x =1 or x = 4/6
i.e., x = 1 or x = 2/3
From the above examples we can deduce the following algorithm for completing the square.
Algorithm : Let the quadratic equation by
ax^2 + bx + c = 0
Step-1 : Divide each side by 'a'
Step-2 : Rearrange the equation so that term c/a is on the right side.(RHS)
Step-3 : Add [1/2(b/a)]^2 to both sides to make LHS, a perfect square.
Step-4 : Write the LHS as a square and simplify the RHS
Step-5 : Solve it.
*******************************************
Example-6. Find the roots of the equation 5x^2-6x-2=0 by the method of completing the square.
sol) Now we follow the Algorithm
Step-1 : x^2 - 6 x - 2 = 0 (divide both side by 5)
5 5
Step-2 : x^2 - 6 x = 2
5 5
Step-3 : x^2 - 6 x + (3)^2 = 2 + (3)^2
5 5 5 5
[ Adding (3/5)^2 to both sides ]
Step-4 : [ x - 3 ]^2 = 2 + 9
5 5 25
Step-5 : ( x - 3 )^2 = 19
5 25
x - 3 = + √19
5 √25
x = 3 + √19 or x = 3 - √19
5 5 5 5
.^. x = 3 + √19 or x = 3 - √19
5 5
*****************************************
Example-7. Find the roots of 4x^2 +3x +5 = 0 by the method of completing the square.
sol) x^2 + 3 x + 5 = 0 ( dividing by 4)
4 4
x^2 + 3 x = -5
4 4
x^2 + 3 x + (3/8)^2 = -5 + (3/8)^2
4 4
(x + 3 )^2 = -5 + 9
8 4 64
( x + 3 )^2 = -71 < 0
8 64
But ( x + 3 )^2 cannot be negative for any real
8
value x .
So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.
*****************************************
We have solved several examples with the use of the method of 'completing the square'. Now, let us apply this method in standard form of quadratic equation ax^2 +bx+c=0(a=/=0)
Step 1 : Dividing the equation through out by 'a' we get
x^2 + b x + c = 0
a a
Step 2 : x^2 + b x = - c
a a
Step-3 : x^2 + b x + [ 1 * b]^2 = - c + [ 1 *b]^2
a 2 a a 2 a
=> x^2 + 2.x. b + [ b ]^2 = - c + [b]^2
2a 2a a 2a
Step-4 : [ x + b]^2 = b^2 - 4ac
2a 4a^2
Step-5 : If b^2-4ac > 0, then by taking the square roots, we get
x + b = + √b^2 - 4 ac
2a 2a
Therefore,
x= -b + √b^2 - 4ac
2a
So, the roots of ax^2 +bx +c=0 are
-b + √b^2 - 4ac and -b-√b^2 -4ac
2a 2a
if b^2 - 4ac > 0.
If b^2 - 4ac < 0, the equation will have no real roots
Thus, if b^2-4ac > 0, then the roots of the quadratic equation ax^2 +bx+c=0 are given by
-b + √b^2-4ac
2a
This formula for finding the roots of a quadratic equation is knows as the quadratic formula.
*****************************************
Example-8. Solve Q.2(!) of Exercise 5.1 by using quadratic formula.
sol) Let the breadth of the plot be x meters.
Then the length is (2x+1) meters.
Since area of rectangular plot is 528 m^2
We can write
x(2x+1) = 528,
i.e, 2x^2 +x-528=0
This is in the form of ax^2 + bx+c = 0,
where a =2, b = 1, c = -528
So, the Quadratic formula gives us the solution as
x = -1 + √1 + 4(2)(528)
4
=> -1 + √4225
4
=> -1 + 65
4
x = 64/4 or x = -66/4
x = 16 or x = -33/2
Since x cannot be negative.
So, the breadth of the plot is 16 meters and hence the length of the plot is (2x+1)=33m.
You should verify that these values satisfy the conditions of the problem.
*****************************************
Example-9. Find two consecutive odd positive integers, sum of whose squares is 290.
sol) Let first off positive integers be x. Then, the second integer will be x+2. According to the question.
x^2 + (x+2)^2 = 290
x^2 + x^2 + 4x + 4 = 290
2x^2 + 4x - 286 = 0
x^2 + 2x - 143 = 0
which is a quadratic equation in x.
Using the quadratic formula
x = -b + √b^2 - 4ac
2
we get,
x = -2 + 4 + 572
2
x = -2 + √576
2
x = -2 + 24
2
x = 11 or x = -13
But x is given to be an odd positive integer.
Therefore, x =/= -13, x = 11.
Thus, the two consecutive odd integers are 11 and (x+2) = 11 + 2 = 13.
Check : 11^2 + 13^2 = 121 + 169 = 290 .
*****************************************
Example-10. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square meters more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12m. Find its length and breadth.
sol) Let the breadth of the rectangular park be x m.
So, its length = (x +3) m.
Therefore, the area of the rectangular park = x(x+3) m^2 = (x^2 + 3x) m^2
Now, base of the isosceles triangle = x m.
Therefore,
its area = 1/2 * x * 12 = 6 x m^2
According to our requirements.
x^2 + 3x = 6x +4
x^2 -3x -4 = 0
Using the quadratic formula, we get
x = 3 + √25 = 3 + 5 = 4 or -1
2 2
But x =/= -1
Therefore, x = 4
So, the breadth of the park= 4m and
its length will be
x+3 = 4 +3 = 7 m
Verification :
Area of rectangular park = 28 m^2
area of triangular park = 24 m^2 = (28-4)m^2
*****************************************
Example-11. Find the roots of the following quadratic equations, if they exist, using the quadratic formula.
(!) x^2 + 4x + 5 = 0
sol) Here a = 1, b = 4 , c = 5.
So, b^2 - 4ac = 16 - 20 = -4 <0
Since the square of a real number cannot be negative, therefore
√b^2 - 4ac will not have any real value.
So, there are no real roots for the given equation.
*****************************************
(!!) 2x^2- 2√2 x + 1 = 0
sol) Here a = 2 , b = -2, c = 1
So, b^2 - 4ac = 8-8 = 0
therefore,
x = 2√2 + √0 = √2 + 0 i.e., x = 1
4 2 √2
So, the roots are 1, 1
√2 √2
*****************************************
Example-12. Find the roots of the following equations :
(!) x + 1 = 3, x=/=0
x
sol) Multiplying whole by x, we get
x^2 + 1 = 3x
i.e., x^2 - 3x + 1= 0, which is a quadratic equation.
Here, a = 1, b = -3, c = 1
So, b^2 - 4ac = 9-4=5>0
Therefore,
x = 3 + √5
2
So, the roots are
3 + √5 and 3 - √5
2 2
*****************************************
(!!) 1 - 1 = 3, x =/=0, 2
x x- 2
sol) As x=/=0,2, multiplying the equation by x(x-2), we get
(x-2)-x= 3x(x-2)
=> 3x^2 -6x
So, the given equation reduces to 3x^2-6x+2=0, which is a quadratic equation.
Here, a =3, b = -6, c =2.
So, b^2 - 4ac = 36-24 =12 > 0
Therefore,
x = 6 + √12
6
=> 6 + 2√3
6
=> 3 + √3
3
So, the roots are
3 + √3 and 3 -√3
3 3
*****************************************
Example-13. A motor boat whose speed is 18 km/h in still water. It takes 1 hour more to go 24km upstream than to return downstream to the same spot. Find the speed of the stream.
sol) Let the speed of the stream be x km/h
Therefore, the speed of the boat upstream
=(18-x) km/h and the speed of the boat downstream = (18 +x) km/h.
The time taken to go upstream
= distance
speed
= 24 hours
18-x
Similarly, the time taken to go downstream
= 24 hours
18+x
According to the question,
24 - 24 = 1
18-x 18+x
i.e., 24(18+x) - 24(18-x) = (18-x)(18+x)
i.e., x^2 + 48x -324 = 0
Using the quadratic formula, we get
x = -48 + √(48)^2 + 1296
2
=> -48 + √3600
2
= -48 + 60 or -54
2
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = -54.
Therefore, x = 6 gives the speed of the stream as 6 km/h.
*****************************************
Exercise - 5.3
1) Find the roots of the following quadratic equations, if they exist, by the method of completing the square.
(1) 2x^2 + x - 4 = 0
sol) Given : 2x^2 + x - 4 = 0 { ax^2 + bx +c ) ----- (1)
where a = 2 , b = 1, c = -4
Now we follow the Algorithm(Steps) :-
Step 1 : Divide each side by "a" here a = 2
=> 2x^2 + x - 4 = 0 ( Dividing both sides by "2")
2 2 2
=> x^2 + x - 2 = 0
2
Step 2 : Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + x = 2
2
Step 3 :- Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
from (eq. (1)) = 2x^2 + x - 4 = 0 ,where b = 1 and a = 2
we ca get "b" and "a" value
i.e [ 1/2 ( 1 /2)]^2 => [ 1/4] ^2
Now , x^2 + x + ( 1 /4) ^2 = 2 + (1/4)^2
2
( Adding (1/4)^2 on both sides)
Step 4 : (x + 1/4)^2 = 2 + 1
16
Step 5 : (x + 1/4) ^2 = 2(16) + 1
16
=> (x + 1/4)^2 = 32 +1
16
=> (x + 1/4) ^2 = 33
16
transfer ^2 to right side 33/16 becomes _/(33/16)
=> x + 1 = + _/(33/16)
4
=> x = _/33 - 1
4 4
=> x = _/33 - 1
4
OR
=> x = - _/33 - 1
4 4
=> x = -- ( _/33 - 1 )
4
the root of the eq are
x = _/33 - 1 & x = - ( _/33 - 1 )
4 4
************************************************************
2) 4x^2 + 4_/3x + 3 = 0
Sol) Given : 4x^2 + 4_/3x + 3 = 0 {ax^2 +bx +c]
Where a = 4, b = 4_/3, c = 3
Step 1:
: 4x^2 + 4_/3 + 3 = 0 ( divide by "a" => a = 4)
4 4 4
=> x^2 + _/3 + 3 = 0
4
Step 2 :
Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + _/3 = - 3
4
Step 3 :
Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
=> [ 1/2 (4_/3 ) ]^2
4
=> [ 1/2 (_/3) ] ^2
=> [ _/3 ] ^2
2
Now,
=> x^2 + _/3x + (_/3)^2 = - 3 + (_/3)^2
2 4 2
Step4 :
=> (x + _/3) ^2 = - 3 + 3
2 4 4
Step 5 :
=> (x + _/3) ^2 = 0
2 4
=> (x + _/3) = + ( 0 ) ^2
2
=> (x + _/3) = 0
2
=) x = - _/3
2
OR
=> x + _/3 = - 0
2
Consider a Kho-Kho court of dimension Legth(29m) * 16m(Breadth).This is to be a rectangular enclosure of area 558 m^2.
They want to leave space of equal width all around the court for the spectators.
What would be the width of the space for spectators? would it be enough?
Suppose the width of the space be 'x' meter. So from the figure
length of the plot would b (29+2x) meter.
Breadth of the rectangular plot would be (16+2x) m
.^., area of the rectangular plot
will be = length * breadth
= (29+2x) * (16 +2x)
Since the area of the plot is = 558 m^2
.^. (29+2x)(16+2x) = 558
.^. 4x^2 + 90x + 464 = 558
=> 4x^2 + 90x - 94 = 0 (dividing by 2)
=> 2x^2 + 45x - 47 = 0 ---(1)
The value of 'x' from the above equation will give the possible width of the space for spectators.
Let us consider another example :
Rani has a square metal sheet. She removed squares of side 9 cm, from each corner of this sheet. Of the remaining sheet, she turned up the sides to form an open box. The capacity of the box is 144 cc. Can we find out the dimensions of the metal sheet?
Suppose the side of the square piece of metal sheet be 'x' cm.
Then, the dimensions of the box are
9 cm. * (x-18) cm. * (x-18) cm.
Since volume of the box is 144 cc
9(x-18)(x-18) = 144
(x-18)^2 = 16
x^2 - 36x + 308 = 0
So, the side 'x' of the metal sheet will satisfy the equation.
x^2 - 36x + 308 ---------(2)
Let us observe the L.H.S of equation (1) and (2)
Are they quadratic polynomials?
Since, the LHS of the above equations are quadratic polynomials they are called quadratic equations.
(*) Quadratic Equations
A quadratic equation in the variable 'x' is an equation of the form ax^2 +bx =c = 0, where a,b,c are real numbers and a=/=0.
Example: 2x^2 + x - 300 = 0 is quadratic equation.
Similarly,
2x^2 -3x +1 =0
4x-3x^2+2 = 0 and
1 - x^2 + 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x)=0, where p(x) is polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation.
That is, ax^2+bx+c=0, a=/=0 is called standard form of a quadratic equation and
y = ax^2+bx+c is called a quadratic equation.
There are various uses o Quadratic functions. Some of them are :-
1) When the rocket is fired upward, then the height of the rocket is defined by a ' quadratic equation.'
2) Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses and orbits of the celestial objects are defined by the quadratic equations.
3) The path of a projectile is defined by quadratic function.
4) When the breaks are applied to a vehicle, the stopping distance is calculated by using quadratic equation.
Example-1. Represent the following situations mathematically:
(!) Raju and Rajendar together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles now they have is 124. We would like to find out how many marbles they had previously.
sol) Let the number of marbles Raju had be 'x'
Then the number of marbles
Rajendar had = 45-x
The number of marbles left with Raju, when he lost 5 marbles = x- 5
The number of marbles left with rajendar, when he lost 5 marbles= (45-x)-5
=> 40-x
Therefore, their product = (x-5)(40-x)
=> 40x - x^2 - 200 +5x
=> -x^2 +45x -200 =124(Given product =124)
=> x^2 -45x +324 = 0 (multiply -ve sign)
Therefore, the number of marbles raju had 'x', satisfies the quadratic equation
x^2 - 45x +324 = 0
Which is the required representation of the problem mathematically.
(!!) The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the other two sides is 5 cm. We would like to find out the length of the two sides?
sol)
Let the length of smaller side be 'x' cm.
Then length of larger side = (x+5) cm.
Given length of hypotenuse = 25 cm.
In a right angle triangle we know that
(hypotenuse)^2 = (side)^2 + (side)2
So,
(25)^2 = x^2 + (x+5)^2
625 = x^2 + x^2 +25 +10x
2x^2 + 10x - 600 = 0
x^2 +5x - 300 = 0
Value of 'x' from the above equation will give the possible value of length of sides of the given right angled triangle.
*****************************************
Example-2. Check whether the following are quadratic equations.
(!) (x-2)^2 + 1 = 2x - 3
sol) LHS = (x-2)^2 + 1
=> x^2 -4x +4 +1
=> x^2 -4x +5
.^., (x-2)^2 +1 = 2x-3 can be written as
x^2 - 4x + 5 = 2x - 3
i.e, x^2 - 6x + 8 = 0
It is in the form of = ax^2 + bx + c = 0
Therefore, the given equation is a quadratic equation.
***************************************
(!!) x(x+1) +8 = (x+2)(x-2)
sol) LHS = x(x+1) + 8 = x^2 +x+8
RHS = (x+2)(x-2) = x^2 - 4
Therefore,
x^2 + x + 8 = x^2 - 4
x^2 +x+8-x^2 +4 = 0
i.e, x + 12= 0
It is not in the form = ax^2 +bx+c = 0
Therefore, the given equation is not a quadratic equation.
****************************************
(!!!) x(2x+3) = x^2 + 1
sol) LHS = 2x^2 + 3x
So, x(2x+3) = x^2 +1 can be written as
2x^2 + 3x = x^2 + 1
Therefore, we get
x^2 + 3x - 1 = 0
It is in the form of= ax^2 +bx+c = 0
So, the given equation is a quadratic equation.
*****************************************
(!V) (x+2)^3 = x^3 - 4
sol) LHS = (x+2)^2 * (x+2)
=> (x^2 + 4x + 4 )* (x+2)
=> x^3 + 2x^2 + 4x^2 + 8x + 4x + 8
=> x^3 + 6x^2 + 12x + 8
Therefore,(x+2)^3 = x^3 -4 can be written as
x^3 + 6x^2 + 12x + 8 = x^3 - 4
i.e, 6x^2 + 12x + 12 = 0
or
x^2 + 2x + 2 = 0
It is in the form of = ax^2 + bx + c = 0
So, the given equation is a quadratic equation,
Remark : In (!!) above, the given equation appears to be a quadratic equation, but is not a quadratic equation.
In (!V) above, the given equation appears to be a cubic equation( an equation of degree 3 ) and not a quadratic equation.
But it turns out to be a quadratic equation.As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.
*****************************************
Exercise - 5.1
1) check whether the following are quadratic equations
(1) (x +1) ^2 = 2 (x -3)
sol)
(x +1) ^2 => (a+b)^2
x^2 + 1^2 + 2 *x*1 => a^2 + b^2 + 2ab
Now,
x^2 + 1 + 2x = 2(x - 3)
x^2 + 1 + 2x = 2x - 6
x^2 + 1 +
x^2 + 7 = 0
We can write,
x^2 + 0x + 7 = 0 { a^2 + bx + c form]
where a = 1 , b = 0, c = 7
Hence, it is quadratic equation.
****************************************************************
2) x^2 - 2x = (-2) (3 -x)
sol) x^2 - 2x = (-2 * 3) - (-2* x)
=> x^2 - 2x = -6 +2x
=> x^2 - 2x +6 - 2x = 0
=> x^2 -4x + 6 = 0 { ax^2 + bx + c form}
where a = 1, b = -4, c = 6
Hence, it is quadratic equation.
*************************************************************
3) (x - 2) (x +1) = (x - 1) (x +3)
sol) x* ( x +1) -2 *(x + 1) = x ( x+3) -1 ( x +3)
=>
=> x - 2x - 2 = 3x - x - 3
=> -x - 2 = 2x -3
=> -x - 2x -2 + 3
=> -3x + 1 = 0 {not equal to =/= ax^2 + bx + c }
Hence, is it not a quadratic equation.
********************************************************
4) (x - 3) ( 2x +1) = x (x+5)
sol) x ( 2x + 1) - 3 (2x +1) = x (x+5)
=>
=> x^2 + x - 6x -3 = 5x
=> x^2 - 5x - 3 = 5x
=> x^2 - 5x - 5x - 3 = 0
=> x^2 - 10x - 3 = 0 { ax^2 + bx + c = 0 form}
where a = 1, b = -10, c = -3
Hence its a quadratic equation.
**********************************************************
5) (2x - 1 ) ( x - 3) = (x +5 ) (x - 1)
sol) 2x ( x - 3) -1 (x - 3) = x ( x - 1) + 5 (x - 1)
=>
=> x^2 - 6x -x + 3 = -x + 5x - 5
=> x^2 - 7x + 3 = 4x - 5
=>x^2 -7x - 4x +3 + 5 = 0
=>x^2 - 11x + 8 = 0 { ax^2 + bx + c form]
where a = 1, b = -11, c = 8
Hence its a quadratic equation.
************************************************************
6) x^2 + 3x + 1 = (x - 2) ^2
sol) x^2 + 3x + 1 = x^2 + 2^2 - 2 * x * 2
******************************************
compare (x-2)^2 with ( a-b)^2 = a^2 + b^2 - 2ab
*******************************************
=>
=> 3x + 1 = 4 - 4x
=> 3x + 1 - 4 + 4x = 0
=> 3x - 3 + 4x = 0
=> 7x - 3 = 0 { =/= ax^2 + bx + c}
Hence, it is not a quadratic solution.
**************************************************************
7) ( x + 2) ^3 = 2x ( x^2 - 1)
sol) x ^3 + 2^3 + 3*x*2 (x+2) = 2x ( x^2 - 1)
**************************************************
compare (x +3)^3 with (a +b)^3 = a^3 + b^3 + 3ab(a+b)
**************************************************
=> x ^3 + 2^3 + 3*x*2 (x+2) = 2x^3 - 2x
=> x^3 + 8 + 6x(x + 2) = 2x^3 - 2x
=> x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x
=> x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 = 0
=> -x^3 + 6x^2 + 14x + 8 = 0 { =/= ax^2 + bx +c}
Hence, it is not a quadratic equation.
*****************************************************************
8) x^3 - 4x^2 - x + 1 = ( x - 2)^3
sol) x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3*x*2(x - 2)
*****************************************************
compare (x - 2)^3 with (a-b)^3 = a^3 - b^3 - 3ab(a - b)
*****************************************************
=> x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x(x -2)
=>
=>-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0
=> 2x^2 - 13x + 9 = 0 { == ax^2 + bx + c}
where a = 2, b = -13 and c = 9
Hence, it is a quadratic equation.
******************************************************************
2) Represent the following situation in the form of quadratic equation.
1) The area of a rectangular plot is 528m.sq . The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
sol) Given that Area = 528m.sq
condition 1:- length of the plot is one more than twice its breadth
let length be = L
Let breadth be = b
So, length (L) = 2b + 1
******************************************
We have , Area of rectangle = length * breadth
******************************************
=> 528 = (2b + 1) * b
=> 528 = 2b^2 + b
=>2b^2 + b - 528 = 0 { ax^2 + bx + c=0}
where a = 2, b = 1, c = -528
Hence, it is quadratic equation.
************************************************************
2) The product of two consecutive positive integers is 306. We need to find the integers.
sol) Let 1st number be = x
Let 2nd number be = x +1
condition :- product of two consecutive integers = 306
=> x * (x+1) = 306
=> x^2 + x = 306
=> x^2 + x - 306 { ax^2 + bx + c form}
where a = 1 , b = 1, c = -360
Hence, it is a quadratic equation.
*****************************************************************
3) Rohan's mother is 26.yr older than him. The product of their ages after 3 .yrs will be 360.yrs.We need to find Rohan's present age.
sol) Let Rohan's age be = x
Condition:- Rohan's mother age will be = x +26
condition 2 :- Product of their ages after 3.YRS will be = 360
After 3 yrs :- Rohan age be = x + 3
Rohan's mother age be = (x+26) + 3 => x +29
product of their ages will be 360 :- (x+3) ( x + 29) = 360
Condition:- Rohan's mother age will be = x +26
condition 2 :- Product of their ages after 3.YRS will be = 360
After 3 yrs :- Rohan age be = x + 3
Rohan's mother age be = (x+26) + 3 => x +29
product of their ages will be 360 :- (x+3) ( x + 29) = 360
=> x ( x +29) + 3 ( x + 29) = 360
=> x^2 + 29x + 3x + 87 = 360
=> x^2 + 29x + 3x + 87 - 360 = 0
=> x^2 + 32x - 273 = 0 { == ax^2 + bx + c form }
where a = 1 , b = 32, c = -273
Hence, it is a quadratic equation.
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4) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hrs more to cover the same distance. We need to find the speed of the train.
sol) Let the speed of train be = x km/hr
Condition 1 :- Train travels 480 km distance at uniform speed.
Distance = 480
Speed = x km/h
We have
Speed = Distance
Time
x = 480
Time
Time = 480
x --------- (1)
Condition 2 :- Speed had been 8km/h less
Distance = 480 km
Speed = ( x - 8) km/hr
3 hr more to cover :- 480 + 3
x
We have
Speed = Distance
Time
(x - 8) = 480
( 480 + 3 )
x
(x- 8) ( 480 + 3 ) = 480 --------------- (2)
x
Solving (2) we get
=> (x - 8) ( 480 + 3 ) = 480
x
=> (x - 8) ( 480 + 3x ) = 480
x
=> (x - 8) ( 480 + 3x) = 480x
=> x ( 480 + 3x) - 8 ( 480 + 3x) = 480x
=>
=> 3x^2 - 24x - 3840 = 0
=> 3 ( x^2 - 8x - 1280) = 0
=> x^2 - 8x - 1280 = 0 { ==ax^2 + bx + c form}
Where a = 1, b= -8, c = -1280
Hence , its quadratic equation
*****************************************************************
(*) Solutions of a Quadratic Equation By factorization:
Consider the quadratic equation 2x^2-3x+1=0. If we replace 'x' by '1'. Then, we get
(2*1^2) -(3*1)+1 = 0 = RHS of the equation.
Since 1 satisfies the equation, we say that 1 is a root of the quadratic equation
2x^2 - 3x + 1= 0.
.^. x=1 is a solution of the quadratic equation.
This also means that 1 is a zero of the quadratic polynomial 2x^2-3x+1.
In general, a real number 𝜶 is called a root of the quadratic equation ax^2 + bx + c = 0.
If a𝜶^2 + b𝜶 + c = 0. We say that x =𝜶 is a solution of the quadratic equation, or 𝜶 satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax^2+bx+c and the roots of the quadratic equation ax^2+bx+c=0 are the same.
We observed that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have atmost two roots.
Example-3. Find the roots of the equation 2x^2-5x+3=0, by factorization.
sol) Let us first split the middle term. recall that if ax^2+bx+c is a quadratic equation polynomial then to split the middle term we have to find two numbers p and q such that
p+q=b and p*q= a*c
So to split the middle term of 2x^2 -5x +3, we have to find two numbers p and q such that
p+q= -5 and p*q = 2*3=6
For this we have to list out all possible pair of factors of 6.
They are (1,6);(-1,-6);(2,3);(-2,-3) . from the list it is clear that the pair (-2,-3) will satisfy our condition p+q=-5 and p*q=6.
The middle term '-5x' can be written as '-2x-3x'.
So, 2x^2 -5x + 3 = 2x^2-2x-3x+3
= 2x(x-1) -3 (x-1) = (2x-3)(x-1)
Now, 2x^2 -5x +3 = 0 can be rewritten as
(2x-3)(x-1) = 0.
So, the values of x for which 2x^2-5x+3=0 are the same for which (2x-3)(x-1) = 0.
i.e, either 2x-3 =0 or x -1=0
Now,
2x-3=0 gives x = 3/2
and
x-1 =0 gives x =1
So, x=3/2 and x=1 are the solutions of the equation.
In other words, 1 and 3/2 are the roots of the equation 2x^2-5x+3=0
Note: We have found the roots of 2x^2-5x+3=0 by factorizing 2x^2-5x+3 into two linear factors and equating each factor to zero.
Example-4. Find the roots of the quadratic equation x - 1/3x = 1/6
sol) 3x^2 - 1 = 1
3x 6
=> 18x^2 - 6 = 3x
=> 18x^2 -3x - 6 = 0 (divide by 3)
=> 6x^2 - x -2 = 0
=> 6x^2 +3x-4x -2 = 0
=> 3x(2x +1) -2(2x+1)
=> (3x-2)(2x+1)
The roots of 6x^2-x-2=0 are the values of x for which (3x-2)(2x+1) = 0
Therefore, 3x-2=0 and 2x+1=0
i.e, x = 2/3 or x = -1/2
Therefore, the roots of 6x^2-x-2 =0 are 2/3 and -1/2
We verify the roots, by checking that 2/3 and -1/2 satisfy 6x^2-x-2=0.
*****************************************
Example-5. Find the width of the space for spectators in kho-kho ground as we discussed above
sol) We found that if the width of the space for spectators is x m., then x satisfies the equation 2x^2 +45x-47=0. Applying the factorization method we write this equation as :-
2x^2 - 2x + 47x - 47 = 0
2x(x-1) + 47(x-1) = 0
i.e, (x-1)(2x+47) = 0
So, the roots of the given equation are x = 1 or x = -47/2. Since 'x' is the width of space of the spectators it cannot be negative.
Thus, the width is 1 m.
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Exercise- 5.2
1) Find the roots of the following quadratic equations by factorisation ;
1) x ^2 = 3x - 10 = 0
sol) x^2 - 2x - 5x - 10 = 0
=> x ( x +2) - 5 ( x + 2) = 0
=> (x - 5) (x +2) = 0
=> x = 5 or x = -2
Hence, x = 5, -2 are the roots of equation.
****************************************************************
2) 2x^2 + x - 6 = 0
sol) 2x^2 + 4x -3x - 6 = 0
=> 2x ( x + 2) - 3 ( x + 2) = 0
=> (2x - 3) ( x +2) = 0
=> 2x - 3 = 0 (x+2) = 0
=> 2x = 3 x = -2
=> x = 3 x = -2
2
Hence x = 3 / 2 , - 2 are the roots of equation.
*****************************************************************
3) _/2 x^2 + 7x + 5_/2 = 0
sol) _/2x^2 + 2x + 5x + 5_/2 = 0
=> _/2x^2 + (_/2 * _/2)x + 5x + 5_/2 = 0
=> _/2x(x + _/2) + 5(x + -/2) = 0
=> (_/2x + 5) (x + -/2) = 0
=> _/2x + 5 = 0 x + -/2 = 0
=> _/2x = -5 x = - _/2
=> x = -5
_/2
Hence, x = -5 , - _/2 are the roots of equation.
_/2
*********************************************************
4) 2x^2 - x + 1 = 0
8
sol)( 8 * 2x^2) - (8*x ) + 1 = 0
8
=> 16x^2 - 8x + 1 = 0 / 8
=> 16x^2 - 4x - 4x + 1 = 0
=> 4x(4x - 1) -1(4x - 1) = 0
=> (4x - 1) (4x -1) = 0
=> 4x - 1 = 0 4x - 1 = 0
=> 4x = 1 and 4x = 1
Hence, x = 1 /4, 1/4 are the roots of equation.
*********************************************************
5) 100x^2 - 20x + 1 = 0
sol) 100x^2 - 10x - 10x + 1 = 0
=> 10x(10x - 1) -1(10x - 1) = 0
=> (10x-1) (10x - 1) = 0
=> 10x = 1 and 10x = 1
=> x = 1/10 and x = 1 / 10
Hence, x = 1 /10, 1/10 are the roots of equation.
*************************************************************
6) x(x + 4) = 12
sol) x^2 + 4x = 12
=> x^2 + 4x -12 = 0
=> x^2 + 6x - 2x - 12 = 0
=> x(x + 6) -2( x + 6) = 0
=> (x - 2 ) ( x +6) = 0
=> x -2 = 0 and x +6 = 0
=> x = 2 and x = -6
Hence, x = 2, -6 are the roots of equation.
*************************************************************
7) 3x^2 - 5x + 2 = 0
sol) 3x^2 - 3x - 2 x + 2 = 0
=> 3x ( x - 1) -2 ( x - 1) = 0
=> (3x - 2) (x - 1) = 0
=> 3x - 2 = 0 and x - 1 = 0
=> x = 2 / 3 and x = 1
Hence, x = 2 /3, 1 are roots of the euqtion.
***********************************************************
8) x - 3 = 2
x
sol) x*x - 3 = 2
x
=> x^2 - 3 = 2 *x
=> x^2 - 2x - 3 = 0
=> x^2 - 3x + x - 3 = 0
=> x(x - 3) + 1 (x -3) = 0
=> (x - 3)(x + 1) = 0
=> x = 3 and x = -1
Hence, x = 3, -1 are the roots of the equation.
************************************************************
9) 3 (x - 4)^2 - 5(x - 4) = 12
***********************************************
sol) compare (x - 4) ^2 with (a - b )^2 = a^2 +b^2 - 2ab
***********************************************
=> 3 (x^2 + 4^2 - 2*x*4 ) -5x + 20 = 12
=> 3( x^2 + 16 -8x) - 5x + 20 - 12 = 0
=> 3x^2 + 48 - 24x - 5x + 8 = 0
=> 3x^2 - 29x + 56 = 0
=> 3x^2 -21x - 8x + 56 = 0
=> 3x ( x - 7) - 8 ( x -7)
=> (3x - 8)(x - 7) = 0
=> 3x - 8 = 0, and x - 7 = 0
=> x = 8 / 3 and x = 7
.^. x = 8/3, 7 are the roots of the equation.
************************************************************
2) Find two number whose sum is 27 and product is 182
sol) let the 1st number be = x
let the 2nd number be = y
Given :- sum of two number = 27
=> x + y = 27 ---- (1)
Given :- product of two number = 182
=> x * y = 182 ------ (2)
using eq.1.. we get y = 27 - x
substitute "y" in eq.2 we get
=> x * y = 182
=> x * (27 - x) = 182
=> 27x - x^2 - 182
=> -x^2 + 27x - 182 = 0
=> x^2 - 27x + 182 = 0
=> x^2 - 13x - 14x + 182 = 0
=> x(x - 13) - 14( x - 13) = 0
=> (x - 14) ( x - 13) = 0
=> x = 14 and x = 13
When x = 13
put "x" value in => x + y = 27
=> 13 + y = 27
=> y = 27 - 13
=> y = 14
.^. when 1st number = 13
2nd number = 14
When x = 14
=> x = 27 - 14
=> x = 13
.^. when 1st number = 14
2nd number = 13
*********************************************************
3) Find two consecutive positive integers, sum of whose squares is 613
sol) let 1st number = x
Let 2nd number = x +1
Condition :- sum of squares = 365
.i.e, x^2 + (x + 1) ^2 = 365
*************************************************
compare (x + 1)^2 with (a + b)^2 = a^2 + b^2 + 2ab
*************************************************
=> x^2 + ( x^2 + 1^2 + 2*x*1) = 365
=> 2x^2 + 1 + 2x = 365
=> 2x^2 + 2x - 365 + 1 = 0
=> 2x^2 + 2x - 364 = 0
=> 2 (x^2 + x - 182) = 0
=> x^2 + x - 182 = 0 / 2
=> x^2 + 14x - 13x - 182 = 0
=> x ( x + 14) - 13( x + 14) = 0
=> (x - 13) (x + 14) = 0
=> x = 13 and x = -14
Since its given in question it is positive integers, we can take only " x = 13" .We cannot take negative "x = -14"
When(first no) x = 13
Then 2nd number = x + 1 => 13 + 1 => 14
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4) the altitude of a right traingle is 7cm less than its base . if the hypotenuse is 13cm, find the other two sides.
sol) Let base of the triangle = x
condition : - the altitude(height) is 7cm less than base(i.e, "x")
.^. height = x- 7
Since It is a right angled triangle
using Pythagoras theorem
Formula : - ( Hypotenuse )^2 = ( height) ^2 + (Base) ^2
=> (AC)^2 = (AB)^2 + (BC)^2
Now,We have:-
Hypotenuse(AC) = 13cm ----(1)
Height (AB) = x - 7 ----(2)
Base( BC) = X ----(3)
Substitute (1), (2), (3) value in formula we get,
> (AC)^2 = (AB)^2 + (BC)^2
=> (13)^2 = (x - 7)^2 + x^2
********************************************
Note:- (x-7)^2 is in (a-b)^2 ={ a^2 + b^2 - 2ab} form
********************************************
=> 13 * 13 = (x^2 + 7^2 - 2(x)(7)) + x^2
=> 169 = x^2 + 49 - 14x + x^2
=>x^2 + x^2 - 14x +49 - 169 = 0
=> 2x^2 -14x - 120 = 0
=> 2(x^2 - 7x - 60) = 0
=> x^2 - 7x - 60 = 0
2
=> x^2 + 5x - 12x - 60 = 0
=> x(x + 5) -12(x + 5) = 0
=> (x - 12)(x + 5) = 0
=> x- 12 = 0 and x +5 = 0
=> x = 12 and x = -5
Since, height cannot be in negative unit we take just x = 12
.^. height = x - 7
= 12 - 7
= 5
and Base = 12
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5) A cottage industry produces a certain number of pottery articles in a day. it was observed on a particular day that the cost of production of each article ( in rupees) was 3 more than twice the number of articles produced on that day. if the total cost of production on that day was Rs,90, find the number of articles produced and the cost of each article.
Sol) Given :- total cost of production = 90Rs
Let the number of articles = x
condition:- cost of production of each article( rupees) was 3 more than twice the number of articles produced on that day
i.e, cost of article = twice the number of articles (2x) + three more (3)
= 2x + 3
We know
Total cost of production = number of articles produced * Cost of article
=> 90 = x*(2x+3)
=> 90 = 2x^2 + 3x
=> 2x^2 + 3x - 90 = 0
=> 2x^2 + 15x - 12x - 90 = 0
=> x ( 2x + 15) -6(2x + 15) = 0
=> (x - 6) (2x + 15) = 0
=> x-6 = 0 and 2x + 15 = 0
=> x = 6 and 2x = -15 => x = -15 / 2
But cost cannot be in negative . thus we take x = 6
.^. Number of articles = 6
and cost of articles = 2x + 1
=2(6) + 1
= 13
.^. Cost of articles = 13rs
****************************************************
6) Find the dimensions of a rectangle whose perimeter is 28mts.and whose area is 40.sq.mt
sol) let length of rectangle be = L
Let breadth of rectangle be = b
Given :- Perimeter is 28mts.
Formula :- perimeter of rectangle (28) = 2(L + b)
=> 2(L + b ) = 28
=> L + b = 28
2
=> L + b = 14
=> b = 14 - L -----------(1)
Given :- Area of rectangle = 40.sq.mt
Formula:- Area of rectangle(40) = L * b
=> 40 = L * b
=> 40 = L * (14 - L)
=> 40 = 14L - L^2
=> 40 - 14l + L^2 = 0
=> L^2 - 14L + 40 = 0
=> l^2 - 10L - 4L + 40 = 0
=> L(L - 10) -4( L - 10) = 0
=> ( L - 4) ( L - 10)
=> L - 4 = 0 or L - 10 = 0
=> l = 4 or L = 10
Substitute "L = 10" in "b= 14 - L"
We get,
=> b = 14 - 10
=> b = 4
Hence , length = 10m and breadth = 4m.
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7) The base of a triangle is 4cm longer than its altitude. if the area of the traingle is 48.sq.cm then find its base and altitude.
sol) Let the altitude be = y
given :- Base is 4cm longer than its altitude
i.e, base = 4 + altitude
= 4 + y
We have area of traingle formula :-
Formula :-
Area of traingle( 48) = 1/2 * base * altitude( height)
=> 48 = 1 * ( 4 + y) * y
2
=> 48 * 2 = 4y + y^2
=> y^2 + 4y - 96 = 0
=> y^2 + 12y - 4y - 96 = 0
=> y(y + 12) -8(y + 12) = 0
=> (y - 8) ( y + 12) = 0
=> y - 8 = 0 or y + 12 = 0
=> y = 8 or y = -12
As Altitude(height) cannot have negative value we ignore y = -12
Now, when y = 8
base = 4 + height
= 4 + 8
= 12
Hence, base = 12 and altitude = 8cm
**************************************************************
8) Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km apart find the average speed of each train.
sol) Let the speed of 1st train be = x km/hr
condition :- 1st train travels 5 km/hr faster than 2nd train.
.^. Speed of 2nd train be = (x - 5) km/hr
condition 2:- After 2 hr they are 50km apart
The second train travels from O to B.
Given that AB = 50 km
It is clear that triangle OAB is a right triangle.
using Pythagoras Theorem, We have,
AB^2 = OA^2 + OB^2 ------ (1)
We know that distance = speed * time
=> OA = x * 2 = 2x -----(2)
similarly, we have,
=> OB = 2 (x - 5) ---------(3)
Substitute (2) and (3) in (1) we get
Given distance between them (AB) = 50 km
AB^2 = OA^2 + OB^2
(50)^2 = (2x)^2 + [2(x-5)]^2
2500 =4x^2 + 4 *{ (x-5)]^2
2500 = 4x^2 + 4 (x^2 + 5^2 - 2 * x *5 )
2500 = 4x^2 + 4 ( x^2 + 25 - 10x)
2500 = 4x^2 + 4x^2 + 100 - 40x
2500 = 8x^2 - 40x + 100
=> 8x^2 - 40x + 100 - 2500 = 0
=> 8x^2 - 40x - 2400 = 0
=>8 ( x^2 - 5x - 300) = 0
=> x^2 - 5x - 300 = 0
8
=> x^2 - 5x - 300 = 0
=> x^2 - 20x + 15x - 300 = 0
=> x(x - 20) +15(x - 20) = 0
=> (x + 15) (x - 20) = 0
=> x + 15 = 0 or x - 20 = 0
=> x = -15 or x = 20
Speed cannot be negative. ignore x = -15
Thus, the speed of the first train is = 20 km/hr
Speed of 2nd train (x -5) = 20 - 5 = 15 km/hr
****************************************************************
9).. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was 1600rs. How many boys are there in the class?
sol) Let number of girls be = x
Let number of boys be = y
Total number of student( x + y) = 60
x + y = 60 ----- (1)
condition :- each boy(x) contributed rupees equal to the number of girls(y) and each girl (y) contributed rupees equal to the number of boys(x)
Total money collected = 1600
xy + yx = 1600
=> 2xy = 1600
=> xy = 1600 / 2
=> xy = 800 -------(2)
From eq,(1) we get, y = 60 -x
substitute (y = 60 -x) in (eq. 2)
=> xy = 800
=> x * (60 - x) = 800
=> 60x - x^2 = 800
=> x^2 - 60x + 800 = 0
=> x^2 - 20x - 40x + 800 = 0
=> x(x - 20) -40(x - 20) = 0
=> (x - 40)(x - 20) = 0
=> x - 40 = 0 or x - 20 =0
=> x = 40 or x = 20
if the number of girls is 20, then the number of boys is 40.
And if the number of girls is 40, Then the number of boys is 20
**************************************************************
10) A motor boat heads upstream a distance of 24km on a river whose current is running at 3 km per hour. The trip up and back takes 6hrs. Assuming that the motor boat maintained a constant speed, what was its speed?
sol) let the speed of the boat be = x km/hr
Speed of the boat upstream = (x - 3) km/hr
Speed of the boat downstream = ( x + 3) km/hr
We have formula :- Time = Distance
Speed
Time taken to go upstream = Distance
Speed.
= 24
x - 3
similarly time taken to downstream = Distance
Speed
= 24
x + 3
Condition :- It took total 6hrs
24 + 24 = 6
x - 3 x + 3
=> 24 ( x + 3) + 24 ( x - 3) = 6 (x - 3) (x +3)
=> 24x +
=> 48x = 6(x^2 - 9)
=> 48x = 6x^2 - 54
=> 6x^2 - 48x - 54 = 0
=> 6 (x^2 - 8x - 9) = 0
=> x^2 - 8x - 9 = 0
6
=> x^2 - 8x - 9 = 0
=> x^2 -9x +x - 9 = 0
=> x(x - 9) + 1 ( x - 9) = 0
=> (x + 1)(x - 9) = 0
=> x+ 1 = 0 or x - 9 = 0
=> x = -1 or x = 9
Since "x" is speed of the stream, it cannot be negative. Ignore x = -1
.^. the speed of the stream is = 9 km/hr
*****************************************************************
(*) Solution of a quadratic equation by completing the square
Is method of factorization applicable to all types of quadratic equation?
let us try to solve x^2+4x-4=0 by factorization method
To solve the given equation x^2 +4x-4=0 by factorization method.
we have to find 'p' and 'q' such that
p+q= 4 and
p*q = -4
But is it not possible. So by factorization method we cannot solve the given equation.
Therefore, we shall study another method.
Consider the following situation
The product of Sunita's age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?
To answer this,
let her present age(in years) be x years.
Age before two year = x-2 &
age after four years = x +4
then the product of both the ages is (x-2)(x+4)
Therefore, (x-2)(x+4) = 2x + 1
i.e., x^2 + 2x -8 = 2x + 1
i.e., x^2 - 9 = 0
So, Sunita's present age satisfies the quadratic equation x^2-9=0
We can write this as x^2= 9.
Taking square roots, we get x =3 or x=-3.
Since the age is a positive number, x = 3.
So, Sunita's present age is 3 years.
Now consider another equation (x+2)^2 -9=0.
To solve it, we can write it as (x+2)^2 = 9
Taking square roots, we get
x+2 = 3 or x + 2 = -3
Therefore x = 1 or x = -5
So, the roots of the equation
(x+2)^2 -9 =0 are 1 and -5.
In both the examples above, the term containing x is completely a square, and we found the roots easily by taking the square roots.
But, what happens if we are asked to solve the equation x^2+4x-4=0. And it cannot be solved by factorization also.
So, we now introduce the method of completing the square. The idea behind this method is to adjust the left side of the quadratic equation so that it becomes a perfect square.
The process is a s follows.
x^2 + 4x - 4 = 0
=> x^2 + 4x = 4
=> x^2 + 2x .2 = 4
Now, the LHS is in the form of a^2 + 2ab. If we add b^2 it becomes as a^2 + 2ab = b^2 which is perfect square. So, by adding b^2 = 2^2 =4 to both sides we get,
x^2 + 2x. 2 + 2^2 = 4 + 4
=> (x+2)^2 = 8
=> x + 2 = +√8
=> x = -2 + 2√2
Now consider the equation 3x^2 -5x +2 =0.
Note that the coefficient of x^2 is not 1.
So we divide the entire equation by 3 so that the coefficient of x^2 is 1
.^. x^2 - 5 x + 2 = 0
3 3
=> x^2 - 5 x = -2
3 3
=> x^2 - 2.x.5 = -2
6 3
=> x^2 - 2.x.5 + (5 )^2 = -2 + (5)^2
6 (6)^2 3 (6)^2
( add (5/6)^2 both side )
( x - 5 )^2 = -2 + 25
6 3 36
(a-b)^2 form
(x - 5 )^2 = (12 * -2)+(25*1)
6 36
(x - 5)^2 = -24 +25
6 36
(x - 5 )^2 = 1
6 36
x - 5 = + 1
6 6
So, x = 5 + 1 or x = 5 - 1
6 6 6 6
Therefore, x =1 or x = 4/6
i.e., x = 1 or x = 2/3
From the above examples we can deduce the following algorithm for completing the square.
Algorithm : Let the quadratic equation by
ax^2 + bx + c = 0
Step-1 : Divide each side by 'a'
Step-2 : Rearrange the equation so that term c/a is on the right side.(RHS)
Step-3 : Add [1/2(b/a)]^2 to both sides to make LHS, a perfect square.
Step-4 : Write the LHS as a square and simplify the RHS
Step-5 : Solve it.
*******************************************
Example-6. Find the roots of the equation 5x^2-6x-2=0 by the method of completing the square.
sol) Now we follow the Algorithm
Step-1 : x^2 - 6 x - 2 = 0 (divide both side by 5)
5 5
Step-2 : x^2 - 6 x = 2
5 5
Step-3 : x^2 - 6 x + (3)^2 = 2 + (3)^2
5 5 5 5
[ Adding (3/5)^2 to both sides ]
Step-4 : [ x - 3 ]^2 = 2 + 9
5 5 25
Step-5 : ( x - 3 )^2 = 19
5 25
x - 3 = + √19
5 √25
x = 3 + √19 or x = 3 - √19
5 5 5 5
.^. x = 3 + √19 or x = 3 - √19
5 5
*****************************************
Example-7. Find the roots of 4x^2 +3x +5 = 0 by the method of completing the square.
sol) x^2 + 3 x + 5 = 0 ( dividing by 4)
4 4
x^2 + 3 x = -5
4 4
x^2 + 3 x + (3/8)^2 = -5 + (3/8)^2
4 4
(x + 3 )^2 = -5 + 9
8 4 64
( x + 3 )^2 = -71 < 0
8 64
But ( x + 3 )^2 cannot be negative for any real
8
value x .
So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.
*****************************************
We have solved several examples with the use of the method of 'completing the square'. Now, let us apply this method in standard form of quadratic equation ax^2 +bx+c=0(a=/=0)
Step 1 : Dividing the equation through out by 'a' we get
x^2 + b x + c = 0
a a
Step 2 : x^2 + b x = - c
a a
Step-3 : x^2 + b x + [ 1 * b]^2 = - c + [ 1 *b]^2
a 2 a a 2 a
=> x^2 + 2.x. b + [ b ]^2 = - c + [b]^2
2a 2a a 2a
Step-4 : [ x + b]^2 = b^2 - 4ac
2a 4a^2
Step-5 : If b^2-4ac > 0, then by taking the square roots, we get
x + b = + √b^2 - 4 ac
2a 2a
Therefore,
x= -b + √b^2 - 4ac
2a
So, the roots of ax^2 +bx +c=0 are
-b + √b^2 - 4ac and -b-√b^2 -4ac
2a 2a
if b^2 - 4ac > 0.
If b^2 - 4ac < 0, the equation will have no real roots
Thus, if b^2-4ac > 0, then the roots of the quadratic equation ax^2 +bx+c=0 are given by
-b + √b^2-4ac
2a
This formula for finding the roots of a quadratic equation is knows as the quadratic formula.
*****************************************
Example-8. Solve Q.2(!) of Exercise 5.1 by using quadratic formula.
sol) Let the breadth of the plot be x meters.
Then the length is (2x+1) meters.
Since area of rectangular plot is 528 m^2
We can write
x(2x+1) = 528,
i.e, 2x^2 +x-528=0
This is in the form of ax^2 + bx+c = 0,
where a =2, b = 1, c = -528
So, the Quadratic formula gives us the solution as
x = -1 + √1 + 4(2)(528)
4
=> -1 + √4225
4
=> -1 + 65
4
x = 64/4 or x = -66/4
x = 16 or x = -33/2
Since x cannot be negative.
So, the breadth of the plot is 16 meters and hence the length of the plot is (2x+1)=33m.
You should verify that these values satisfy the conditions of the problem.
*****************************************
Example-9. Find two consecutive odd positive integers, sum of whose squares is 290.
sol) Let first off positive integers be x. Then, the second integer will be x+2. According to the question.
x^2 + (x+2)^2 = 290
x^2 + x^2 + 4x + 4 = 290
2x^2 + 4x - 286 = 0
x^2 + 2x - 143 = 0
which is a quadratic equation in x.
Using the quadratic formula
x = -b + √b^2 - 4ac
2
we get,
x = -2 + 4 + 572
2
x = -2 + √576
2
x = -2 + 24
2
x = 11 or x = -13
But x is given to be an odd positive integer.
Therefore, x =/= -13, x = 11.
Thus, the two consecutive odd integers are 11 and (x+2) = 11 + 2 = 13.
Check : 11^2 + 13^2 = 121 + 169 = 290 .
*****************************************
Example-10. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square meters more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12m. Find its length and breadth.
sol) Let the breadth of the rectangular park be x m.
So, its length = (x +3) m.
Therefore, the area of the rectangular park = x(x+3) m^2 = (x^2 + 3x) m^2
Now, base of the isosceles triangle = x m.
Therefore,
its area = 1/2 * x * 12 = 6 x m^2
According to our requirements.
x^2 + 3x = 6x +4
x^2 -3x -4 = 0
Using the quadratic formula, we get
x = 3 + √25 = 3 + 5 = 4 or -1
2 2
But x =/= -1
Therefore, x = 4
So, the breadth of the park= 4m and
its length will be
x+3 = 4 +3 = 7 m
Verification :
Area of rectangular park = 28 m^2
area of triangular park = 24 m^2 = (28-4)m^2
*****************************************
Example-11. Find the roots of the following quadratic equations, if they exist, using the quadratic formula.
(!) x^2 + 4x + 5 = 0
sol) Here a = 1, b = 4 , c = 5.
So, b^2 - 4ac = 16 - 20 = -4 <0
Since the square of a real number cannot be negative, therefore
√b^2 - 4ac will not have any real value.
So, there are no real roots for the given equation.
*****************************************
(!!) 2x^2- 2√2 x + 1 = 0
sol) Here a = 2 , b = -2, c = 1
So, b^2 - 4ac = 8-8 = 0
therefore,
x = 2√2 + √0 = √2 + 0 i.e., x = 1
4 2 √2
So, the roots are 1, 1
√2 √2
*****************************************
Example-12. Find the roots of the following equations :
(!) x + 1 = 3, x=/=0
x
sol) Multiplying whole by x, we get
x^2 + 1 = 3x
i.e., x^2 - 3x + 1= 0, which is a quadratic equation.
Here, a = 1, b = -3, c = 1
So, b^2 - 4ac = 9-4=5>0
Therefore,
x = 3 + √5
2
So, the roots are
3 + √5 and 3 - √5
2 2
*****************************************
(!!) 1 - 1 = 3, x =/=0, 2
x x- 2
sol) As x=/=0,2, multiplying the equation by x(x-2), we get
(x-2)-x= 3x(x-2)
=> 3x^2 -6x
So, the given equation reduces to 3x^2-6x+2=0, which is a quadratic equation.
Here, a =3, b = -6, c =2.
So, b^2 - 4ac = 36-24 =12 > 0
Therefore,
x = 6 + √12
6
=> 6 + 2√3
6
=> 3 + √3
3
So, the roots are
3 + √3 and 3 -√3
3 3
*****************************************
Example-13. A motor boat whose speed is 18 km/h in still water. It takes 1 hour more to go 24km upstream than to return downstream to the same spot. Find the speed of the stream.
sol) Let the speed of the stream be x km/h
Therefore, the speed of the boat upstream
=(18-x) km/h and the speed of the boat downstream = (18 +x) km/h.
The time taken to go upstream
= distance
speed
= 24 hours
18-x
Similarly, the time taken to go downstream
= 24 hours
18+x
According to the question,
24 - 24 = 1
18-x 18+x
i.e., 24(18+x) - 24(18-x) = (18-x)(18+x)
i.e., x^2 + 48x -324 = 0
Using the quadratic formula, we get
x = -48 + √(48)^2 + 1296
2
=> -48 + √3600
2
= -48 + 60 or -54
2
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = -54.
Therefore, x = 6 gives the speed of the stream as 6 km/h.
*****************************************
Exercise - 5.3
1) Find the roots of the following quadratic equations, if they exist, by the method of completing the square.
(1) 2x^2 + x - 4 = 0
sol) Given : 2x^2 + x - 4 = 0 { ax^2 + bx +c ) ----- (1)
where a = 2 , b = 1, c = -4
Now we follow the Algorithm(Steps) :-
Step 1 : Divide each side by "a" here a = 2
=>
2
Step 2 : Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + x = 2
2
Step 3 :- Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
from (eq. (1)) = 2x^2 + x - 4 = 0 ,where b = 1 and a = 2
we ca get "b" and "a" value
i.e [ 1/2 ( 1 /2)]^2 => [ 1/4] ^2
Now , x^2 + x + ( 1 /4) ^2 = 2 + (1/4)^2
2
( Adding (1/4)^2 on both sides)
Step 4 : (x + 1/4)^2 = 2 + 1
16
Step 5 : (x + 1/4) ^2 = 2(16) + 1
16
=> (x + 1/4)^2 = 32 +1
16
=> (x + 1/4) ^2 = 33
16
transfer ^2 to right side 33/16 becomes _/(33/16)
=> x + 1 = + _/(33/16)
4
=> x = _/33 - 1
4 4
=> x = _/33 - 1
4
OR
=> x = - _/33 - 1
4 4
=> x = -- ( _/33 - 1 )
4
the root of the eq are
x = _/33 - 1 & x = - ( _/33 - 1 )
4 4
************************************************************
2) 4x^2 + 4_/3x + 3 = 0
Sol) Given : 4x^2 + 4_/3x + 3 = 0 {ax^2 +bx +c]
Where a = 4, b = 4_/3, c = 3
Step 1:
:
=> x^2 + _/3 + 3 = 0
4
Step 2 :
Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + _/3 = - 3
4
Step 3 :
Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
=> [ 1/2 (
=> [ _/3 ] ^2
2
Now,
=> x^2 + _/3x + (_/3)^2 = - 3 + (_/3)^2
2 4 2
Step4 :
=> (x + _/3) ^2 = - 3 + 3
2 4 4
Step 5 :
=> (x + _/3) ^2 = 0
2 4
=> (x + _/3) = + ( 0 ) ^2
2
=> (x + _/3) = 0
2
=) x = - _/3
2
OR
=> x + _/3 = - 0
2
=> x = - _/3
2
.^. Roots of eq are
x = - _/3 & x = - _/3
x = - _/3 & x = - _/3
2 2
*********************************************************
3) 5x^2 - 7x - 6 = 0
sol) Given : 5x^2 - 7x - 6 = 0 { ax^2 +bx +c}
where a = 5, b = -7, c = -6
Step1 :
divide by a = 5, we get
=> 5x^2 - 7x - 6 = 0
5 5 5
=> x^2 - 7x - 6
5 5
Step 2 :
Rearrange the equation so that constant term c /a is on right side (RHS)
=> { 1/2 ( -7 )^2 ]
5
=> ( -7 ) ^2
10
Now add on both side we get
=> x^2 - 7x + ( -7 )^2 = 6 + ( -7)^2
5 10 5 10
Step 4:
=> ( x - 7 )^2 = 6 + 49
10 5 100
=> ( x - 7 )^2 = (6*100 )+ (49*5)
10 5*100
=> ( x - 7 )^2 = 600 + 245
10 500
=> ( x - 7 )^2 = 845
10 500
=> ( x - 7 )^2 = 5(169)
10 5 (100)
Transfer ^2 on right side 169 / 100 become _/(169/100)
=> (x - 7) = _/(169/100)
10
=>(x - 7) = + 13
10 10
=> x - 7 = 13
10 10
=> x = 13 + 7
10 10
=> x = 20 = 2
10
OR
x = 7 - 13
10 10
x = 7 - 13
10 10
x = -6
10
Hence , 2 & -6 are two roots of the equation.
10
*********************************************************
4) x^2 + 5 = -6x
sol) Given : x^2 + 6x + 5 = 0 {ax^2 + bx + x = 0}
Where a = 1, b = 6, c = 5
Step 1) :-
No need to divide as the co-efficient of "x^2 " is already "1"
Step 2 :-
Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + 6x = -5
Step 3 :-
2 1
*********************************************************
3) 5x^2 - 7x - 6 = 0
sol) Given : 5x^2 - 7x - 6 = 0 { ax^2 +bx +c}
where a = 5, b = -7, c = -6
Step1 :
divide by a = 5, we get
=> 5x^2 - 7x - 6 = 0
5 5 5
=> x^2 - 7x - 6
5 5
Step 2 :
Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 -7x = 6
5 5
Step 3:
Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
=> { 1/2 ( -7 )^2 ]
5
=> ( -7 ) ^2
10
Now add on both side we get
=> x^2 - 7x + ( -7 )^2 = 6 + ( -7)^2
5 10 5 10
Step 4:
=> ( x - 7 )^2 = 6 + 49
10 5 100
=> ( x - 7 )^2 = (6*100 )+ (49*5)
10 5*100
=> ( x - 7 )^2 = 600 + 245
10 500
=> ( x - 7 )^2 = 845
10 500
=> ( x - 7 )^2 =
10
Transfer ^2 on right side 169 / 100 become _/(169/100)
=> (x - 7) = _/(169/100)
10
=>(x - 7) = + 13
10 10
=> x - 7 = 13
10 10
=> x = 13 + 7
10 10
=> x =
OR
x = 7 - 13
10 10
x = 7 - 13
10 10
x = -6
10
Hence , 2 & -6 are two roots of the equation.
10
*********************************************************
4) x^2 + 5 = -6x
sol) Given : x^2 + 6x + 5 = 0 {ax^2 + bx + x = 0}
Where a = 1, b = 6, c = 5
Step 1) :-
No need to divide as the co-efficient of "x^2 " is already "1"
Step 2 :-
Rearrange the equation so that constant term c /a is on right side (RHS)
=> x^2 + 6x = -5
Step 3 :-
Add [ 1/2 (b/a)^2 ] to both sides to make LHS a perfect square
=> { 1 ( 6 ) } ^2
=> { 3 } ^2
now, add (3)^2 on both side we get,
=> x^2 + 6x + 3^2 = -5 + 3^2
=> (x + 3)^2 = -5 + 9
Step 4 :-
=> (x +3) ^2 = 4
=> x + 3 = _/4
=> x + 3 = + 2
=> x = 2 - 3 or x = -2 -3
=> x = -1 or x = -5
Hence, -1, -5 are the roots of the equation.
****************************************************
2) Find the roots of the quadratic equation by applying the quadratic formula?
1) 2x^2 + x - 4 = 0
sol) Given :- 2x^2 + x - 4 =0 ( ax^2 + bx +c=0)
where a = 2, b = 1, c = -4
We have:- A = b^2 - 4ac
D = (1)^2 - 4 (2) (-4)
D = 1 + 32
D = 33 > 0
Since the square of a real number is positive,
.^. _/A will have a real value.
So, the roots of equation are
x = -b + _/A
2a
x = -1 + _/33
2*2
x = -1 + _/33
4
Hence , -1 + _/33 & -1 -_/33 are roots of equation.
4 4
8
2 2
************************************************************
2) 1 - 1 = 11 x =/= -4, 7
x +4 x - 7 30
sol) (x - 7) - (x +4) = 11
(x+4)(x -7) 30
=> x - 7 - x - 4 = 11
(x+4)(x -7) 30
=> -11 = 11
(x + 4)(x - 7) 30
=> -1 = 1
(x +4) (x - 7) 30
=> - 30 = (x +4)(x - 7)
=> x (x - 7) + 4( x -7) = -30
=> x^2 - 7x + 4x - 28 = -30
=> x^2 - 3x - 28 + 30 = 0
=> x^2 - 3x + 2 = 0 (ax^2 + bx + c = 0)
where a = 1, b = -3, c = 2
We have
A = b^2 - 4ac
A = (-3)^2 - 4* 1 * (2)
A = 9 - 8
A = 1
Formula :- x = -b + _/A
2a
x = - (-3) + _/1 or x = -(-3) - _/1
2 * 1 2*1
x= 3 + 1 or x = 3 - 1
2 2
x = 4 or x = 2
2 2
x = 2 or 1 are the roots of the equation.
*********************************************************
4) The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5years from now is 1/3, Find his present age.
Sol) let rehman's present age = x
3yrs ago = x - 3
5yrs from now = x +5
Condition :-
Sum of reciprocal of Rehman's ages 3 yrs ago and 5yrs from
now = 1 / 3
= > 1 + 1 = 1
(x - 3) (x + 5) 3
=> x +5 + x - 3 = 1
(x-3)(x +5) 3
=> 2x + 2 = 1
(x - 3) (x +5) 3
=> 3 * (2x +2) = 1 * (x - 3) ( x + 5)
=> 6x + 6 = x (x + 5) - 3 ( x +5)
=> 6x + 6 = x^2 + 5x -3x -15
=> x^2 + 2x - 15 - 6 - 6x = 0
=> x^2 - 4x - 21 = 0
factorize :-
=> x^2 + 3x - 7x - 21 = 0
=> x ( x + 3) -7(x = 3) = 0
=> (x - 7) ( x + 3) = 0
=> x - 7 = 0 or x + 3 = 0
=> x = 7 or x = -3
But age cannot be negative we ignore x = -3 value
.^. Rehman's current age = 7yrs
*********************************************************
5) In a class test, the sum of Moulika's marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English , the product of her marks would have been 210. Find her Marks in the two subjects.
Sol) Let Moulika's marks in Mathematics = x
Let Moulika's marks in English = y
Given :- Sum of marks in both subject = 30
i.e, x + y = 30
.^. English marks (y) = 30 - x ----- (1)
condition :- If she got 2 marks more in Maths and 3 marks less in english, the product of her marks would have been 210
2 more marks in maths :- x + 2
3 marks less in english : 30 - x - 3
product is 210 => (x + 2) ( 30 - x - 3) = 210
=> (x + 2) ( 27 - x) = 210
=> x(27 -x) +2(27 - x) = 210
=> 27x - x^2 + 54 - 2x = 210
=> -x^2 + 25x + 54 - 210 = 0
=> -x^2 + 25x - 156 = 0
=> x^2 - 25x + 156 = 0
Factorize :-
=> x^2 -12x -13x + 156 = 0
=> x (x - 12) -13(x - 12) = 0
=> ( x -13)(x - 12) = 0
=> x = 13 or x = 12
if Marks in Maths (x = 13)
Then Mars in ENglish = 30 - x => 30 - 13 => 17
if marks in maths (x = 12)
Then marks in English = 30 - x => 30 - 12 = 18
***********************************************************
6) The diagonal of a rectangular field is 60 meters more than the shorter side. if the longer side is 30 meters more than the shooter side, find the sides of the field.
sol) let the shorter side be = x meters
Given :- Longer side is 30 meters more than the shooter side = x + 30 mts
Given :- Diagonal is 60 meters more than shorter side = x + 60 mts
By Pythagoras theorem
AC^2 = (AB)^2 + (BC)^2
(x + 60)^2 = x^2 + (x + 30)^2
x^2 + 3600 + 120x = x^2 + x^2 + 900 + 60x
-x^2 + 120x - 60x + 3600 - 900 = 0
-x^2 + 60x + 2700 = 0
x^2 - 60x - 2700 = 0
factorise :-
x^2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x + 30) ( x - 90) = 0
x = - 30 or x = 90
Since "x" length cannot be negative ignore x = -30
.^. Shorter side of the field (X) = 90 meters
Longer side = X + 30 => 90 + 30 => 120 meters
*********************************************************
7) The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
sol) Let larger number be = x
Given :- Square of the smaller no. is 8 times the larger number
i.e, smaller .no.(y)^2 = 8 * Larger number
=> (y)^2 = 8 * x
=> y^2 = 8x
=> y = _/8x ----- (1)
condition :- Difference of squares of two numbers = 180
=> (x)^2 - (y)^2 = 180
=> x^2 - ( _/8x)^2 = 180
=> x^2 - 8x = 180
=> x^2 - 8x - 180 = 0
Factorise :-
=x^2 - 8x - 180 = 0
=> x^2 + 10x - 18x - 180 = 0
=> x( x + 10) - 18( x + 10) = 0
=> (x - 18) ( x + 10) = 0
=> x = 18 or x = - 10
When larger number (X) = 18
smaller no(y) = _/(8x) => _/ (8 * 18) => _/(144) => _/(12)^2 = 12
********************************************************
8) A train travels 360km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
sol) Let the speed of train be = x km/hr
Uniform speed ;-
Distance = 360 km
Speed = x km/hr
We have
Speed = Distance
Time
x = 360
Time
Time = 360
x
When speed is 5km/hr more :-
Distance = 360km
Speed = ( x + 5) km /hr
Time = 360 - 1 ( One hour less time)
x
We have
Speed = Distance
Time
x + 5 = 360
(360 - 1)
x
(x + 5) ( 360 - 1) = 360 -----(1)
x
solve (1) to get the equation.
(x + 5) ( 360 - x) = 360
x
(x + 5) (360 - x) = 360x
x(360 - x) +5(360 - x) = 360x
360x - x^2 + 1800 - 5x = 360x
-x^2 -5x + 1800 = 0
x^2 + 5x - 1800 = 0
factorise :-
x^2 + 45x - 40x - 1800 = 0
x ( x + 45) - 40 ( x + 45) = 0
(x + 45) ( x - 40) = 0
x = -45 or x = 40
Since speed cannot be in negative ignore x = -45
.^. Speed of train (X) = 40 km /hr
******************************************************
9) Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Sol) Let time taken by smaller tap to fill be = x hours
Volume of tank filled by smaller tap in 1 hour = 1 / x
Given :-Time taken by larger tap = 10 hr less than smaller tap
i.e, x - 10 hours
Volume of tank filled by larger tap in 1 hour = 1
x - 10
Given :- time taken by both taps to fill = 9 3/8 hours
i.e, 9(8) + 3 hours = 75 hours
8 8
Tank filled by smaller tap in 75 hrs = 1 * 75
8 x 8
i.e, 75
8x
tank filled by larger tap in 75 hrs = 1 * 75
8 x - 10 8
=> 75
8(x - 10)
Tank filled by smaller tap + Tank filled by larger tap = 1
=> 75 + 75 = 1
8x 8(x - 10)
=> 75 ( 1 + 1 ) = 1
8 x x -10
=> 1 + 1 = 8
x x -10 75
=> x -10 + x = 8
x(x - 10) 75
=> 2x - 10 = 8
x^2 - 10x 75
=> 75( 2x - 10) = 8(x^2 -10x)
=> 150x - 750 = 8x^2 - 80x
=> 150x - 750 -8x^2 + 80x = 0
=> -8x^2 + 150x + 80x - 750 = 0
=> -8x^2 + 230x - 750 = 0
=> 8x^2 - 230x + 750 = 0 { ax^2 + bx + c = 0}
Where a = 8, b = -230, c = 750
We have
A = b^2 -4ac
A = (-230)^2 - 4*8 * 750
A = 52900 - 24000
A = 28900
Roots of equations :-
x = -b + _/A
2a
x = -(-230) + _/(28900)
2 * 8
x = 230 + _/28900
16
x = 230 + _/( 289 * 100)
16
x = 230 + _/(289) * _/100
16
x = 230 + _/(17)^2 * _/(10)^2
16
x = 230 + 17 * 10
16
x = 230 + 170
16
solving, x = 230 + 170 or x = 230 - 170
16 16
x = 400 or x = 60
16 16
x = 25 or x = 15 / 4
Now,
Time taken by smaller tap x = 25
Then larger tap takes = x - 10
= 25 - 10 = 15 hrs
When x = 15 / 4
Time taken by larger tap = x - 10
=> 15 - 10
4
=> 15 - 40
4
=> - 25
4
Since it's negative .Time can't b negative
************************************************************
10) An express train takes 1 hour less than a passenger train to travel 132km between Mysore and bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express trains is 11 km/h more than that of the passenger train, find the average speed of the two trains.
sol) Let the average speed of passenger train be = x km/hr
passenger train :-
Distance = 132 km
Speed = x
Speed = Distance
Time
x = 132
Time
Time = 132
x
Express train :-
Distance = 132 km
Given :- Speed of express train is 11 km/hr more than passenger
So, Speed = x + 11
Speed = Distance
Time
x + 11 = 132
Time
Time = 132
(x + 11)
Given :- train taken 1 hr less
.^. Time taken by passenger train - Time taken by express = 1 train
132 - 132 = 1
x (x + 11)
132(x + 11) - 132x = 1
x(x + 11)
132x + 1452 - 132x = 1
x^2 + 11x
1452 = 1
x^2 + 11x
1452 = x^2 + 11x
x^2 + 11x - 1452 = 0 { ax^2 = bx + c= 0}
where a = 1, b = 11, c = - 1452
We have
A = b^2 - 4ac
A = (11)^2 - 4 *1 * (-1452)
A = 121 + 5808
A = 5929
So, the roots of equation are
x = -b + _/A
2a
x = -11 + _/ 5929
2* 1
x = - 11 + _/(77)^2
2
x = -11 + 77
2
Solving , we get
x = -11 + 77
2
x = 66 / 2
x = 33
OR
x = -11 - 77
2
x = -88 / 2
x = - 44
Since speed cannot be negative , we ignore x = -44
.^. Average speed of passenger train (X) = 33 km/hr
So, average speed of express train = x + 11 => 33 = 11 = 44 km/hr
*********************************************************
11) Sum of the areas of two squares is 468 m^2. If the difference of their perimeters is 24 m. Find the sides of the two squares.
Sol) Let side of square 1 be = x meters
Perimeter of square 1 = 4 * side => 4x
Given :- Difference of perimeter of squares is 24 m
Perimeter of square 1 - Perimeter of square 2 = 24
4x - Perimeter of square 2 = 24
4x - 24 = perimeter of square 2
4x - 24 = 4* (side of square 2)
4x - 24 = side of square 2
4
4(x - 6) = side of square 2
4
Hence,
side of square 1 is = x
side of square 2 = x - 6
Given :- sum of area of square is 468 m
Area of square 1 + Area of square 2 = 468
(Side of square )^2 + ( Side of square 2 ) ^2 = 468
x^2 + (x - 6)^2 = 468
x^2 + x^2 + 6^2 - 2 (x)(6) = 468
x^2 + x^2 + 36 - 12x = 468
2x^2 - 12x +36 - 468 = 0
2x^2 - 12x - 432 = 0
2(x^2 - 6x - 216) = 0
x^2 - 6x - 216 = 0 / 2
x^2 - 6x - 216 = 0 { ax^2 + bx + c = 0]
where a = 1, b = -6, c = -216
We have :-
A = b^2 - 4ac
A = (-6)^2 - 4 * 1 * ( -216)
A = 36 + 864
A = 900
The roots of equations are :-
x = -b + _/A
2a
x = -(-6) + _/900
2a
x = 6 + 30
2*1
Solving
x = 6 + 30
2*1
x = 36 / 2
x = 18
OR
x = 6 - 30
2*1
x = -24 / 2
x = -12
Since, side cannot be negative, ignore x = -12
.^. side of square 1 = 18m
side of square 2 = x - 6 => 18 - 6 => 12m
*********************************************************
12) A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity 80 m/second. The distance 's' of the ball from the ground after 1 seconds is
S = 96 + 80t - 16t^2. After how many seconds does the ball strike the ground.
Sol) Given :- 16t^2 + 80t + 96 = 0
=> 16t^2 - 80t - 96 = 0
=> 16( t^2 - 5t - 6) = 0
=> t^2 - 5t - 6 = 0 /16
=> t^2 - 5t - 6 = 0
factorise :-
t^2 - 6t + t - 6 = 0
t(t - 6) + 1(t -6) = 0
(t +1) ( t - 6) = 0
t = -1 or t = 6
But time cannot be negative so ignore t = -1
Hence t = 6 seconds
*********************************************************
13) If a polygon of 'n' sides has 1/2 n (n -3) diagonals. How many sides will a polygon having 65 diagonals? is there a polygon with 50 diagonals?
sol) 1 n(n - 3) = 65
2
=> n(n -3) = 65 * 2
=> n^2 - 3n = 130
=> n^2 - 3n - 130 = 0
=> n^2 - 13n + 10n - 130 = 0
=> n(n - 13) + 10 ( n - 13) = 0
=> (n + 10) ( n - 13) = 0
=> n = -10 or n = 13
But sides cannot be negative
Hence n = 13
.^. Polygon having 65 diagonals will have 13 sides.
2) 1 n(n - 3) = 50
2
=> n(n - 3) = 50 * 2
=> n^2 - 3n = 100
=> n^2 - 3n - 100 = 0 (ax^2 + bx + c= 0)
where a = 1, b = -3, c = -100
we have,
A = b^2 - 4ac
A = (-3)^2 - 4 * 1 * (-100)
A = 9 + 400
A = 409
Now,
x = -b + _/A
2a
x = -(-3) + _/409
2*1
x = 3 + _/409
2
But the side should be an integer, So a polygon cannot have 50 diagonals.
*************************************************************
(*) Nature of Roots :
In the previous section, we have seen that the roots of the equation ax^2 +bx +c = 0 are given by
x = -b + √b^2 - 4ac
2a
Now let us try to understand the nature of roots.
Remember that zeros are those points where value of polynomial becomes zero or we can say that the curve of quadratic polynomial cuts the x-axis
Similarly, roots of a quadratic equation are those points where the curve cuts the X-axis.
Case-1 : If b^2 - 4ac > 0
We get two distinct roots
-b + √b^2 - 4ac , -b -√b^2 - 4ac
2a 2a
In such case if we draw graph for the given quadratic we get the following figures.
Figure shows that the curve of the quadratic equation cuts the X-axis at two distinct points
Case-2 : If b^2 - 4ac = 0
x = -b + 0
2a
So, x = - b , -b
2a 2a
Figure shows that the curve of the quadratic equation touching X-axis at one point.
Case-3 : b^2 - 4ac < 0
There are no real roots. Roots are imaginary.
In this case graph neither intersects nor touches the X-axis at all. So, there are no real roots.
Since b^2-4ac determines whether the quadratic equation ax^2 + bx + c = 0 has real roots or not, b^2 -4ac is called discriminant of the quadratic equation.
So, a quadratic equation ax^2 +bx+c = 0 has
(!) two distinct real roots, if b^2 - 4ac> 0
(!!) two equal real roots, if b^2 -4ac=0.
(!!!) no real roots, if b^2-4ac < 0.
Let us consider some examples.
Example-14. Find the discriminant of the quadratic equation 2x^2 - 4x + 3 = 0, and hence find the nature of its roots.
sol) The given equation is in the form of
ax^2 + bx + c = 0,
where a =2, b=-4 and c = 3. Therefore, the discriminant.
b^2-4ac = (-4)^2 - 4(2)(3)
=> 16- 24 = -8 < 0
So, the given equation has no real roots.
******************************************
Example-15. A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters.
Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
sol) let us first draw the diagram.
Let P be the required location of the pole.
Let the distance of the pole from the gate B be x m
i.e, BP= x m.
Now the difference of the distances of the pole from the two gates = AP - BP
( or BP-AP) = 7 m.
Therefore, AP = ( x +7) m.
Now, AB = 13 m, and since AB is a diameter.
∠APB = 90 deg
Therefore,
AP^2 + PB^2 = AB^2 ( By Pythagoras theorem)
(x+7)^2 + x^2 = (13)^2
x^2 + 14x + 49 + x^2 = 169
2x^2 + 14x - 120 = 0
So, the distance 'x' of the pole from gate B satisfies the equation.
x^2 + 7x - 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this is so or not,
let us consider its discriminant. The discriminant is
b^2 - 4ac = 7^2 - 4 ( 1) (-60)
=> 289 > 0
So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park..
Solving the quadratic equation
x^2 + 7x - 60 = 0, by the quadratic formula, we get
x = -7 + √289
2
= -7 + 17
2
Therefore, x = 5 or -12
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = -12 will have to be ignored .
So, x = 5
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.
******************************************
Example-16. Find the discriminant of the equation 3x^2 - 2x + 1 = 0 and hence find the
3
name of its roots. Find them, if they are real.
sol) Here a = 3, b = -2 and c = 1/3
Therefore, discriminant
b^2 -4ac= (-2)^2 - 4(3)(1/3)
=> 4 - 4 = 0.
Hence, the given quadratic equation has two equal roots.
The roots are
-b , -b
2a 2a
i.e., 2 , 2
6 6
i.e., 1 , 1
3 3
*****************************************
Exercise - 5.4
1) Find the nature of the roots of the following quadratic equations. If real roots exists, find them :
1) 2x^2 - 3x + 5 = 0
sol) 2x^2 - 3x + 5 = 0 { ax^2 + bx + c = 0}
where a = 2, b = -3, c = 5
We have,
A = b^2 - 4ac
A = (-3)^2 - 4* 2 * 5
A = 9 - 40
A = - 31
Since A < 0
Hence, there are no real roots
******************************************************
2) 3x^2 - 4_/3x + 4 = 0
sol) Where a = 3, b = -4_/3, c = 4
We have,
A = b^2 - 4ac
A = (-4_/3)^2 - 4 * 3*4
A = 16*3 - 48
A = 48 - 48
A = 0
Since , A = 0 , we have two equal real roots.
using quadratic formula :-
x = - b + _/A
2a
x = - ( -4_/3) + _/0
2 * 3
x = 4_/3 + 0
6
x = 2_/3
3
Hence, x = 2 _/3 & x = 2_/3 are the roots of the equation.
3 3
*****************************************************
3) 2x^2 - 6x + 3 = 0
sol) Where a = 2, b = -6, c = 3
We have
A = b^2 - 4ac
A = (-6)^2 - 4 * 3 *3
A = 36 - 24
A = 12
Since A > 0 , There are 2 distinct real roots.
Using quadratic formula :-
x = -b + _/A
2a
x = -(-6) + _/12
2*2
x = 6 + _/12
4
x = 6 + _/4 * _/3
4
x = 6 + 2_/3
4
x = 2 (3 + _/3)
4
x = 3 + _/3
2
.^. the root of the equation are x = 3 + _/3 & x = 3 - _/3
2 2
*************************************************************
2) Find the values of "k" for each of the following quadratic equations, so that they have two equal roots.
1) 2x^2 + kx + 3 = 0
sol) Given = 2x^2 + kx + 3 = 0 { ax^2 + bx + c= 0 }
Where a = 2, b = k, c = 3
Given : they are equal roots.
b^2 - 4ac = 0
k^2 - 4*2*3 = 0
k^2 - 24 = 0
k^2 = 24
k = + _/24
k = + _/6*4
k = + 2_/6
*****************************************************
2) kx (x - 2) + 6 = 0
sol) kx^2 - 2kx + 6 = 0 { ax^2 + bx + c = 0}
Where a = k, b = -2k, c = 6
Given :- Equation have equal roots (i.e D = 0)
b^2 - 4ac = 0
(-2k)^2 - 4 * k * 6 = 0
4k^2 - 24k = 0
4k( k - 6) = 0
4k = 0 or k - 6 = 0
k = 0 or k = 6
For (k = 0)
The equation becomes
=> kx^2 - 2kx + 6 = 0
=> 0(x^2) - 2(0)x + 6 = 0
=> 0 - 0 + 6 = 0
=> 6 = 0
Which is not correct, so with, k = 0 is not possible
For(k = 6)
The equation becomes
=> kx^2 - 2kx + 6 = 0
=> 6x^2 - 2(6)x + 6 = 0
=> 6(x^2 -2x + 1) = 0
=> x^2 - 2x + 1 = 0
=> x^2 -x -x + 1 = 0
=> x(x - 1) -1 (x - 1) = 0
=> (x - 1) = 0 or (x - 1 ) = 0
=> x = 1 or x = 1
.^. we got equal roots when (k = 6)
******************************************************
3) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m.sq? If so, find its length and breadth.
sol) Yes it's possible to design rectangular mango grove.
condition :-
Given :- Area = 800m.sq
Let breadth be = x
Length is twice its breadth (y) = 2x
Formula : Area = Length * breadth
800 = x * 2x
800 = 2x^2
x^2 = 800 / 2
x^2 = 400
x = + _/400
x = + 20
we got x = 20 or x = -20
Since breadth cannot be in negative ,,ignore x = -20 value
Thus, Breadth of mango grove (X) = 20m
Length of mango grove = 2x = 2(20) = 40m
***********************************************************
4) The sum of the ages of two friends is 20yrs. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present age.
sol) Let age of 1st friend be = x
Given :- Sum of the ages = 20yrs
i.e, x + y = 20
=> 2nd friend age (y) = 20 -x
Four years ago
1st friend age = x - 4
2nd friend age = (20 - x) - 4 = 16 -x
Product of their ages 4 yrs ago = 48
(x - 4) ( 16 - x) = 48
x(16 -x) -4(16 -x ) = 48
16x - x^2 - 64 + 4x = 48
-x^2 + 20x - 64 - 48 = 0
-x^2 + 20x - 112 = 0
x^2 - 20x + 112 = 0 { ax^2 + bx + c = 0}
Where a = 1, b = -20, c = 112
We have
D = b^2 - 4ac
D = (-20)^2 - 4* 1* 112
D = 400 - 448
D = -48
Since D < 0
The equation has no real roots
Hence, the given situation is not possible
*********************************************************
5) Is it possible to design a rectangular park of perimeter 80m, and area 400m.sq. If so, find its length and breadth.
sol) Let length be = x
Given :- perimeter = 80m
We know perimeter of rectangle = 2(L + b)
=> 2(X + b) = 80
=> X + b = 80/2
=> b = 40 -x
.^. Breadth be(b) = 40 -x
Given :- Area = 400m.sq
We know area of rectangle = L * B
Area = L * b
400 = x(40 -x)
400 = 40x - x^2
x^2 - 40x + 400 = 0
Factorize :-
x^2 - 20x - 20x + 400 = 0
x(x - 20) -20(x - 20) = 0
(x - 20) ( x - 20) = 0
so, x = 20 is the solution.
.^. Length(X) = 20m
Breadth = 40 - x => 40 -20 = 20m
**************************************
(*) Optional Exercise
1) A two digit number is such that the product of the digits is 8. When 18 is added to the number they interchange their places. determine the number.
sol) Let the ten's digit be 'x' and unit's digit b 'y'
.^. Number = 10x +y
Given: product of the digits = 8
xy = 8
y = 8 --------(1)
x
Given: when '18' is added to number they interchange their places.
10x +y+18 = 10y +x
9x - 9y +18=0
x - y + 2= 0-----(2)
put(1) in (2)
x - (8) +2 =0
x
x^2 - 8 + 2x = 0
x^2 +2x -8 = 0
x^2+ 4x-2x-8 = 0
x(x+4)-2(x+4)=0
x = 2 or x=-4
rejecting x=-4 , we have x = 2
put x=2 in (1)
y= 2 =1
2
Thus, the required number is
10x +y = 0
10*2 + 1 = 21 is the required number
*****************************************
2) A piece of wire 8m. in length, cut into two pieces, and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2m^2?
( Hint : x+y=8, (x/4)^2 +(y/4)^2 = 2
=> (x/4)^2 + ((8-x)/4))^2 = 2 ]
sol) Using Hint we can solve directly
=> x^2 + (64+x^2-16x) =2
16 16
=> x^2 + 64 + x^2 - 16x = 32
=> 2x^2 - 16x +32 = 0
=> x^2 - 8x + 16 = 0
=> x^2 -4x -4x +16 = 0
=> x(x-4) -4(x-4)= 0
=> x = 4
Length of one of the piece of wire = 4m
Given : x +y = 8
4 + y = 8
y = 4
length of another piece of wire=4m
.^.,the cut in the wire should be made exactly middle of the wire.
****************************************
3). vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take vinay to complete the job by himself.
sol) praveena can complete the job = x days
Per day praveena can do the job = 1
x
Given:- Vinay can complete the job in 5 days less than praveen
i.e, v=x-5 ---(1)
per day vinay can do the job = 1
(x-5)
Working together they can do the job in 6 days.
It means per day they do the job together
= 1
6
Total work in one day by vinay and praveen
1 + 1 = 1
x (x-5) 6
6[(x-5) + x] = x(x-5)
6[ 2x-5] = x^2 - 5x
12x - 30 = x^2 - 5x
x^2 - 17x +30 = 0
x^2 - 15x -2x + 30 = 0
x(x-15) -2(x -15) = 0
x = 2 or x = 15
But its given that vinay can do the job 5 days faster than praveen
put x = 2 in (1)
v = 2 - 5 = -3 days
put x = 15 in (1)
v = 15 -5 = 10
we take v = 10 as work cannot b done in negative(-3 days)
.^., Vinay can complete work in 10 days
*****************************************
4) The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2*(16/21), find the fraction.
sol) Given: Denominator of a fraction is one(+1) more than twice(2) the numerator(x).
Let Numerator = x
denominator = one more twice(2x) the numerator= 2x + 1
original Fraction = x
2x + 1
mixed fraction = 2(16/21) = 42+16 = 58
21 21
Given: sum of fraction and its reciprocal = 58
21
so, the following balanced equation will be formed.
fraction + reciprocal
x + 2x +1 = 58
2x+1 x 21
x^2 + 4x^2 + 4x + 1 = 58
x(2x+1) 21
21(x^2 + 4x^2 + 4x +1) = 58(2x^2 + x)
21(5x^2 + 4x + 1) = 116x^2 + 58x
105x^2 + 84x + 21 = 116x^2 + 58x
11x^2 - 26x - 21 = 0
11x^2 - 33x +7x - 21 = 0
11x(x - 3) + 7(x -3) = 0
(11x+7) = 0 (x-3) = 0
x = -7/11 or x = 3
we will take x = 3
Now
denominator = 2x +1 = 2(3) +1
=> 6 +1 = 7
Thus, fraction is = 3/7
2) Find the roots of the quadratic equation by applying the quadratic formula?
1) 2x^2 + x - 4 = 0
sol) Given :- 2x^2 + x - 4 =0 ( ax^2 + bx +c=0)
where a = 2, b = 1, c = -4
We have:- A = b^2 - 4ac
D = (1)^2 - 4 (2) (-4)
D = 1 + 32
D = 33 > 0
Since the square of a real number is positive,
.^. _/A will have a real value.
So, the roots of equation are
x = -b + _/A
2a
x = -1 + _/33
2*2
x = -1 + _/33
4
Hence , -1 + _/33 & -1 -_/33 are roots of equation.
4 4
*********************************************************
2) 4x^2 + 4_/3x + 3 = 0
sol) Given : 4x^2 + 4_/3x + 3 = 0 ( ax^2 + bx + c = 0)
Where a = 4, b = 4_/3, c = 3
We have,
A = b^2 - 4ac
A = (4_/3)^2 - 4 * 4 * 3
A = (4)^2 * (_/3) ^2 - 48
A = 16 * 3 - 48
A = 48 - 48
A = 0
Formula :- x = -b + _/A
2a
x = - (4_/3) + _/0
2 *4
x = - (4_/3)
x = - (_/3)
2
.^. x = - _/3 & x = _/3 are the roots of the equation.
2 2
************************************************************
3) 5x^2 - 7x - 6 = 0
sol) Here a = 5, b = -7, c = -6
We have
A = b^2 - 4ac
A = (-7)^2 - 4 *5*(-6)
A = 49 +120
A = 169
Formula = x = -b + _/A
sol) Here a = 5, b = -7, c = -6
We have
A = b^2 - 4ac
A = (-7)^2 - 4 *5*(-6)
A = 49 +120
A = 169
Formula = x = -b + _/A
2a
x = - (-7) + _/169 (13 *13 = 169)
(2*5)
x = 7 + 13
10
Now
x = 7 + 13 & 7 - 13
10 10
x = 20 & -6
10 10
x = 2 & -6 /10 are roots of the equation.
****************************************************
x = - (-7) + _/169 (13 *13 = 169)
(2*5)
x = 7 + 13
10
Now
x = 7 + 13 & 7 - 13
10 10
x =
x = 2 & -6 /10 are roots of the equation.
****************************************************
4) x^2 + 5 = -6x
sol) x^2 + 6x + 5 = 0 { ax^2 + bx + c = 0}
where a = 1 , b = 6, c = 5
sol) x^2 + 6x + 5 = 0 { ax^2 + bx + c = 0}
where a = 1 , b = 6, c = 5
we have n
A = b^2 - 4ac
A = (6)^2 - 4 * 1* 5
A = 36 - 20
A = 16
Formula :- x = -b + _/A
2a
x = -6 + _/16
2*1
x = -6 + 4 or x = -6 -4
2 2
x = -2 or x = -10
2 2
x = -1 & -5 are roots of the equation
*********************************************************
3. Find the roots of the following equation.
1) x - 1 = 3 x =/=0
x
sol) x*x - 1 = 3
x
=> x^2 - 1 = 3
x
=> x^2 - 1 = 3x
=> x^2 - 3x - 1 = 0 { ax^2 + bx + c }
Where a = 1, b = -3, c = -1
We have
A = b^2 - 4ac
A = (-3)^2 - 4 * 1 * (-1)
A = 9 + 4
A = 13
Formula :- x = -b + _/A
2a
x = -(-3) + _/13 or x = -(-3) - _/13
2* 1 2 *1
x = 3 + _/13 or x = 3 - _/13
2 2
Hence, roots of this equation is x = 3 + _/13 or x = 3 - _/13A = b^2 - 4ac
A = (6)^2 - 4 * 1* 5
A = 36 - 20
A = 16
Formula :- x = -b + _/A
2a
x = -6 + _/16
2*1
x = -6 + 4 or x = -6 -4
2 2
x = -2 or x = -10
2 2
x = -1 & -5 are roots of the equation
*********************************************************
3. Find the roots of the following equation.
1) x - 1 = 3 x =/=0
x
sol) x*x - 1 = 3
x
=> x^2 - 1 = 3
x
=> x^2 - 1 = 3x
=> x^2 - 3x - 1 = 0 { ax^2 + bx + c }
Where a = 1, b = -3, c = -1
We have
A = b^2 - 4ac
A = (-3)^2 - 4 * 1 * (-1)
A = 9 + 4
A = 13
Formula :- x = -b + _/A
2a
x = -(-3) + _/13 or x = -(-3) - _/13
2* 1 2 *1
x = 3 + _/13 or x = 3 - _/13
2 2
2 2
************************************************************
2) 1 - 1 = 11 x =/= -4, 7
x +4 x - 7 30
sol) (x - 7) - (x +4) = 11
(x+4)(x -7) 30
=> x - 7 - x - 4 = 11
(x+4)(x -7) 30
=> -
(x + 4)(x - 7) 30
=> -1 = 1
(x +4) (x - 7) 30
=> - 30 = (x +4)(x - 7)
=> x (x - 7) + 4( x -7) = -30
=> x^2 - 7x + 4x - 28 = -30
=> x^2 - 3x - 28 + 30 = 0
=> x^2 - 3x + 2 = 0 (ax^2 + bx + c = 0)
where a = 1, b = -3, c = 2
We have
A = b^2 - 4ac
A = (-3)^2 - 4* 1 * (2)
A = 9 - 8
A = 1
Formula :- x = -b + _/A
2a
x = - (-3) + _/1 or x = -(-3) - _/1
2 * 1 2*1
x= 3 + 1 or x = 3 - 1
2 2
x =
x = 2 or 1 are the roots of the equation.
*********************************************************
4) The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5years from now is 1/3, Find his present age.
Sol) let rehman's present age = x
3yrs ago = x - 3
5yrs from now = x +5
Condition :-
Sum of reciprocal of Rehman's ages 3 yrs ago and 5yrs from
now = 1 / 3
= > 1 + 1 = 1
(x - 3) (x + 5) 3
=> x +5 + x - 3 = 1
(x-3)(x +5) 3
=> 2x + 2 = 1
(x - 3) (x +5) 3
=> 3 * (2x +2) = 1 * (x - 3) ( x + 5)
=> 6x + 6 = x (x + 5) - 3 ( x +5)
=> 6x + 6 = x^2 + 5x -3x -15
=> x^2 + 2x - 15 - 6 - 6x = 0
=> x^2 - 4x - 21 = 0
factorize :-
=> x^2 + 3x - 7x - 21 = 0
=> x ( x + 3) -7(x = 3) = 0
=> (x - 7) ( x + 3) = 0
=> x - 7 = 0 or x + 3 = 0
=> x = 7 or x = -3
But age cannot be negative we ignore x = -3 value
.^. Rehman's current age = 7yrs
*********************************************************
5) In a class test, the sum of Moulika's marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English , the product of her marks would have been 210. Find her Marks in the two subjects.
Sol) Let Moulika's marks in Mathematics = x
Let Moulika's marks in English = y
Given :- Sum of marks in both subject = 30
i.e, x + y = 30
.^. English marks (y) = 30 - x ----- (1)
condition :- If she got 2 marks more in Maths and 3 marks less in english, the product of her marks would have been 210
2 more marks in maths :- x + 2
3 marks less in english : 30 - x - 3
product is 210 => (x + 2) ( 30 - x - 3) = 210
=> (x + 2) ( 27 - x) = 210
=> x(27 -x) +2(27 - x) = 210
=> 27x - x^2 + 54 - 2x = 210
=> -x^2 + 25x + 54 - 210 = 0
=> -x^2 + 25x - 156 = 0
=> x^2 - 25x + 156 = 0
Factorize :-
=> x^2 -12x -13x + 156 = 0
=> x (x - 12) -13(x - 12) = 0
=> ( x -13)(x - 12) = 0
=> x = 13 or x = 12
if Marks in Maths (x = 13)
Then Mars in ENglish = 30 - x => 30 - 13 => 17
if marks in maths (x = 12)
Then marks in English = 30 - x => 30 - 12 = 18
***********************************************************
6) The diagonal of a rectangular field is 60 meters more than the shorter side. if the longer side is 30 meters more than the shooter side, find the sides of the field.
sol) let the shorter side be = x meters
Given :- Longer side is 30 meters more than the shooter side = x + 30 mts
Given :- Diagonal is 60 meters more than shorter side = x + 60 mts
By Pythagoras theorem
AC^2 = (AB)^2 + (BC)^2
(x + 60)^2 = x^2 + (x + 30)^2
-x^2 + 120x - 60x + 3600 - 900 = 0
-x^2 + 60x + 2700 = 0
x^2 - 60x - 2700 = 0
factorise :-
x^2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x + 30) ( x - 90) = 0
x = - 30 or x = 90
Since "x" length cannot be negative ignore x = -30
.^. Shorter side of the field (X) = 90 meters
Longer side = X + 30 => 90 + 30 => 120 meters
*********************************************************
7) The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
sol) Let larger number be = x
Given :- Square of the smaller no. is 8 times the larger number
i.e, smaller .no.(y)^2 = 8 * Larger number
=> (y)^2 = 8 * x
=> y^2 = 8x
=> y = _/8x ----- (1)
condition :- Difference of squares of two numbers = 180
=> (x)^2 - (y)^2 = 180
=> x^2 - ( _/8x)^2 = 180
=> x^2 - 8x = 180
=> x^2 - 8x - 180 = 0
Factorise :-
=x^2 - 8x - 180 = 0
=> x^2 + 10x - 18x - 180 = 0
=> x( x + 10) - 18( x + 10) = 0
=> (x - 18) ( x + 10) = 0
=> x = 18 or x = - 10
When larger number (X) = 18
smaller no(y) = _/(8x) => _/ (8 * 18) => _/(144) => _/(12)^2 = 12
********************************************************
8) A train travels 360km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
sol) Let the speed of train be = x km/hr
Uniform speed ;-
Distance = 360 km
Speed = x km/hr
We have
Speed = Distance
Time
x = 360
Time
Time = 360
x
When speed is 5km/hr more :-
Distance = 360km
Speed = ( x + 5) km /hr
Time = 360 - 1 ( One hour less time)
x
We have
Speed = Distance
Time
x + 5 = 360
(360 - 1)
x
(x + 5) ( 360 - 1) = 360 -----(1)
x
solve (1) to get the equation.
(x + 5) ( 360 - x) = 360
x
(x + 5) (360 - x) = 360x
x(360 - x) +5(360 - x) = 360x
-x^2 -5x + 1800 = 0
x^2 + 5x - 1800 = 0
factorise :-
x^2 + 45x - 40x - 1800 = 0
x ( x + 45) - 40 ( x + 45) = 0
(x + 45) ( x - 40) = 0
x = -45 or x = 40
Since speed cannot be in negative ignore x = -45
.^. Speed of train (X) = 40 km /hr
******************************************************
9) Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Sol) Let time taken by smaller tap to fill be = x hours
Volume of tank filled by smaller tap in 1 hour = 1 / x
Given :-Time taken by larger tap = 10 hr less than smaller tap
i.e, x - 10 hours
Volume of tank filled by larger tap in 1 hour = 1
x - 10
Given :- time taken by both taps to fill = 9 3/8 hours
i.e, 9(8) + 3 hours = 75 hours
8 8
Tank filled by smaller tap in 75 hrs = 1 * 75
8 x 8
i.e, 75
8x
tank filled by larger tap in 75 hrs = 1 * 75
8 x - 10 8
=> 75
8(x - 10)
Tank filled by smaller tap + Tank filled by larger tap = 1
=> 75 + 75 = 1
8x 8(x - 10)
=> 75 ( 1 + 1 ) = 1
8 x x -10
=> 1 + 1 = 8
x x -10 75
=> x -10 + x = 8
x(x - 10) 75
=> 2x - 10 = 8
x^2 - 10x 75
=> 75( 2x - 10) = 8(x^2 -10x)
=> 150x - 750 = 8x^2 - 80x
=> 150x - 750 -8x^2 + 80x = 0
=> -8x^2 + 150x + 80x - 750 = 0
=> -8x^2 + 230x - 750 = 0
=> 8x^2 - 230x + 750 = 0 { ax^2 + bx + c = 0}
Where a = 8, b = -230, c = 750
We have
A = b^2 -4ac
A = (-230)^2 - 4*8 * 750
A = 52900 - 24000
A = 28900
Roots of equations :-
x = -b + _/A
2a
x = -(-230) + _/(28900)
2 * 8
x = 230 + _/28900
16
x = 230 + _/( 289 * 100)
16
x = 230 + _/(289) * _/100
16
x = 230 + _/(17)^2 * _/(10)^2
16
x = 230 + 17 * 10
16
x = 230 + 170
16
solving, x = 230 + 170 or x = 230 - 170
16 16
x = 400 or x = 60
16 16
x = 25 or x = 15 / 4
Now,
Time taken by smaller tap x = 25
Then larger tap takes = x - 10
= 25 - 10 = 15 hrs
When x = 15 / 4
Time taken by larger tap = x - 10
=> 15 - 10
4
=> 15 - 40
4
=> - 25
4
Since it's negative .Time can't b negative
************************************************************
10) An express train takes 1 hour less than a passenger train to travel 132km between Mysore and bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express trains is 11 km/h more than that of the passenger train, find the average speed of the two trains.
sol) Let the average speed of passenger train be = x km/hr
passenger train :-
Distance = 132 km
Speed = x
Speed = Distance
Time
x = 132
Time
Time = 132
x
Express train :-
Distance = 132 km
Given :- Speed of express train is 11 km/hr more than passenger
So, Speed = x + 11
Speed = Distance
Time
x + 11 = 132
Time
Time = 132
(x + 11)
Given :- train taken 1 hr less
.^. Time taken by passenger train - Time taken by express = 1 train
132 - 132 = 1
x (x + 11)
132(x + 11) - 132x = 1
x(x + 11)
132x + 1452 - 132x = 1
x^2 + 11x
1452 = 1
x^2 + 11x
1452 = x^2 + 11x
x^2 + 11x - 1452 = 0 { ax^2 = bx + c= 0}
where a = 1, b = 11, c = - 1452
We have
A = b^2 - 4ac
A = (11)^2 - 4 *1 * (-1452)
A = 121 + 5808
A = 5929
So, the roots of equation are
x = -b + _/A
2a
x = -11 + _/ 5929
2* 1
x = - 11 + _/(77)^2
2
x = -11 + 77
2
Solving , we get
x = -11 + 77
2
x = 66 / 2
x = 33
OR
x = -11 - 77
2
x = -88 / 2
x = - 44
Since speed cannot be negative , we ignore x = -44
.^. Average speed of passenger train (X) = 33 km/hr
So, average speed of express train = x + 11 => 33 = 11 = 44 km/hr
*********************************************************
11) Sum of the areas of two squares is 468 m^2. If the difference of their perimeters is 24 m. Find the sides of the two squares.
Sol) Let side of square 1 be = x meters
Perimeter of square 1 = 4 * side => 4x
Given :- Difference of perimeter of squares is 24 m
Perimeter of square 1 - Perimeter of square 2 = 24
4x - Perimeter of square 2 = 24
4x - 24 = perimeter of square 2
4x - 24 = 4* (side of square 2)
4x - 24 = side of square 2
4
Hence,
side of square 1 is = x
side of square 2 = x - 6
Given :- sum of area of square is 468 m
Area of square 1 + Area of square 2 = 468
(Side of square )^2 + ( Side of square 2 ) ^2 = 468
x^2 + (x - 6)^2 = 468
x^2 + x^2 + 6^2 - 2 (x)(6) = 468
x^2 + x^2 + 36 - 12x = 468
2x^2 - 12x +36 - 468 = 0
2x^2 - 12x - 432 = 0
2(x^2 - 6x - 216) = 0
x^2 - 6x - 216 = 0 / 2
x^2 - 6x - 216 = 0 { ax^2 + bx + c = 0]
where a = 1, b = -6, c = -216
We have :-
A = b^2 - 4ac
A = (-6)^2 - 4 * 1 * ( -216)
A = 36 + 864
A = 900
The roots of equations are :-
x = -b + _/A
2a
x = -(-6) + _/900
2a
x = 6 + 30
2*1
Solving
x = 6 + 30
2*1
x = 36 / 2
x = 18
OR
x = 6 - 30
2*1
x = -24 / 2
x = -12
Since, side cannot be negative, ignore x = -12
.^. side of square 1 = 18m
side of square 2 = x - 6 => 18 - 6 => 12m
*********************************************************
12) A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity 80 m/second. The distance 's' of the ball from the ground after 1 seconds is
S = 96 + 80t - 16t^2. After how many seconds does the ball strike the ground.
Sol) Given :- 16t^2 + 80t + 96 = 0
=> 16t^2 - 80t - 96 = 0
=> 16( t^2 - 5t - 6) = 0
=> t^2 - 5t - 6 = 0 /16
=> t^2 - 5t - 6 = 0
factorise :-
t^2 - 6t + t - 6 = 0
t(t - 6) + 1(t -6) = 0
(t +1) ( t - 6) = 0
t = -1 or t = 6
But time cannot be negative so ignore t = -1
Hence t = 6 seconds
*********************************************************
13) If a polygon of 'n' sides has 1/2 n (n -3) diagonals. How many sides will a polygon having 65 diagonals? is there a polygon with 50 diagonals?
sol) 1 n(n - 3) = 65
2
=> n(n -3) = 65 * 2
=> n^2 - 3n = 130
=> n^2 - 3n - 130 = 0
=> n^2 - 13n + 10n - 130 = 0
=> n(n - 13) + 10 ( n - 13) = 0
=> (n + 10) ( n - 13) = 0
=> n = -10 or n = 13
But sides cannot be negative
Hence n = 13
.^. Polygon having 65 diagonals will have 13 sides.
2) 1 n(n - 3) = 50
2
=> n(n - 3) = 50 * 2
=> n^2 - 3n = 100
=> n^2 - 3n - 100 = 0 (ax^2 + bx + c= 0)
where a = 1, b = -3, c = -100
we have,
A = b^2 - 4ac
A = (-3)^2 - 4 * 1 * (-100)
A = 9 + 400
A = 409
Now,
x = -b + _/A
2a
x = -(-3) + _/409
2*1
x = 3 + _/409
2
But the side should be an integer, So a polygon cannot have 50 diagonals.
*************************************************************
(*) Nature of Roots :
In the previous section, we have seen that the roots of the equation ax^2 +bx +c = 0 are given by
x = -b + √b^2 - 4ac
2a
Now let us try to understand the nature of roots.
Remember that zeros are those points where value of polynomial becomes zero or we can say that the curve of quadratic polynomial cuts the x-axis
Similarly, roots of a quadratic equation are those points where the curve cuts the X-axis.
Case-1 : If b^2 - 4ac > 0
We get two distinct roots
-b + √b^2 - 4ac , -b -√b^2 - 4ac
2a 2a
In such case if we draw graph for the given quadratic we get the following figures.
Figure shows that the curve of the quadratic equation cuts the X-axis at two distinct points
Case-2 : If b^2 - 4ac = 0
x = -b + 0
2a
So, x = - b , -b
2a 2a
Figure shows that the curve of the quadratic equation touching X-axis at one point.
Case-3 : b^2 - 4ac < 0
There are no real roots. Roots are imaginary.
In this case graph neither intersects nor touches the X-axis at all. So, there are no real roots.
Since b^2-4ac determines whether the quadratic equation ax^2 + bx + c = 0 has real roots or not, b^2 -4ac is called discriminant of the quadratic equation.
So, a quadratic equation ax^2 +bx+c = 0 has
(!) two distinct real roots, if b^2 - 4ac> 0
(!!) two equal real roots, if b^2 -4ac=0.
(!!!) no real roots, if b^2-4ac < 0.
Let us consider some examples.
Example-14. Find the discriminant of the quadratic equation 2x^2 - 4x + 3 = 0, and hence find the nature of its roots.
sol) The given equation is in the form of
ax^2 + bx + c = 0,
where a =2, b=-4 and c = 3. Therefore, the discriminant.
b^2-4ac = (-4)^2 - 4(2)(3)
=> 16- 24 = -8 < 0
So, the given equation has no real roots.
******************************************
Example-15. A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters.
Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
sol) let us first draw the diagram.
Let P be the required location of the pole.
Let the distance of the pole from the gate B be x m
i.e, BP= x m.
Now the difference of the distances of the pole from the two gates = AP - BP
( or BP-AP) = 7 m.
Therefore, AP = ( x +7) m.
Now, AB = 13 m, and since AB is a diameter.
∠APB = 90 deg
Therefore,
AP^2 + PB^2 = AB^2 ( By Pythagoras theorem)
(x+7)^2 + x^2 = (13)^2
x^2 + 14x + 49 + x^2 = 169
2x^2 + 14x - 120 = 0
So, the distance 'x' of the pole from gate B satisfies the equation.
x^2 + 7x - 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this is so or not,
let us consider its discriminant. The discriminant is
b^2 - 4ac = 7^2 - 4 ( 1) (-60)
=> 289 > 0
So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park..
Solving the quadratic equation
x^2 + 7x - 60 = 0, by the quadratic formula, we get
x = -7 + √289
2
= -7 + 17
2
Therefore, x = 5 or -12
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = -12 will have to be ignored .
So, x = 5
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.
******************************************
Example-16. Find the discriminant of the equation 3x^2 - 2x + 1 = 0 and hence find the
3
name of its roots. Find them, if they are real.
sol) Here a = 3, b = -2 and c = 1/3
Therefore, discriminant
b^2 -4ac= (-2)^2 - 4(3)(1/3)
=> 4 - 4 = 0.
Hence, the given quadratic equation has two equal roots.
The roots are
-b , -b
2a 2a
i.e., 2 , 2
6 6
i.e., 1 , 1
3 3
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Exercise - 5.4
1) Find the nature of the roots of the following quadratic equations. If real roots exists, find them :
1) 2x^2 - 3x + 5 = 0
sol) 2x^2 - 3x + 5 = 0 { ax^2 + bx + c = 0}
where a = 2, b = -3, c = 5
We have,
A = b^2 - 4ac
A = (-3)^2 - 4* 2 * 5
A = 9 - 40
A = - 31
Since A < 0
Hence, there are no real roots
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2) 3x^2 - 4_/3x + 4 = 0
sol) Where a = 3, b = -4_/3, c = 4
We have,
A = b^2 - 4ac
A = (-4_/3)^2 - 4 * 3*4
A = 16*3 - 48
A = 48 - 48
A = 0
Since , A = 0 , we have two equal real roots.
using quadratic formula :-
x = - b + _/A
2a
x = - ( -4_/3) + _/0
2 * 3
x = 4_/3 + 0
6
x = 2_/3
3
Hence, x = 2 _/3 & x = 2_/3 are the roots of the equation.
3 3
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3) 2x^2 - 6x + 3 = 0
sol) Where a = 2, b = -6, c = 3
We have
A = b^2 - 4ac
A = (-6)^2 - 4 * 3 *3
A = 36 - 24
A = 12
Since A > 0 , There are 2 distinct real roots.
Using quadratic formula :-
x = -b + _/A
2a
x = -(-6) + _/12
2*2
x = 6 + _/12
4
x = 6 + _/4 * _/3
4
x = 6 + 2_/3
4
x =
x = 3 + _/3
2
.^. the root of the equation are x = 3 + _/3 & x = 3 - _/3
2 2
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1) 2x^2 + kx + 3 = 0
sol) Given = 2x^2 + kx + 3 = 0 { ax^2 + bx + c= 0 }
Where a = 2, b = k, c = 3
Given : they are equal roots.
b^2 - 4ac = 0
k^2 - 4*2*3 = 0
k^2 - 24 = 0
k^2 = 24
k = + _/24
k = + _/6*4
k = + 2_/6
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2) kx (x - 2) + 6 = 0
sol) kx^2 - 2kx + 6 = 0 { ax^2 + bx + c = 0}
Where a = k, b = -2k, c = 6
Given :- Equation have equal roots (i.e D = 0)
b^2 - 4ac = 0
(-2k)^2 - 4 * k * 6 = 0
4k^2 - 24k = 0
4k( k - 6) = 0
4k = 0 or k - 6 = 0
k = 0 or k = 6
For (k = 0)
The equation becomes
=> kx^2 - 2kx + 6 = 0
=> 0(x^2) - 2(0)x + 6 = 0
=> 0 - 0 + 6 = 0
=> 6 = 0
Which is not correct, so with, k = 0 is not possible
For(k = 6)
The equation becomes
=> kx^2 - 2kx + 6 = 0
=> 6x^2 - 2(6)x + 6 = 0
=> 6(x^2 -2x + 1) = 0
=> x^2 - 2x + 1 = 0
=> x^2 -x -x + 1 = 0
=> x(x - 1) -1 (x - 1) = 0
=> (x - 1) = 0 or (x - 1 ) = 0
=> x = 1 or x = 1
.^. we got equal roots when (k = 6)
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3) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m.sq? If so, find its length and breadth.
sol) Yes it's possible to design rectangular mango grove.
condition :-
Given :- Area = 800m.sq
Let breadth be = x
Length is twice its breadth (y) = 2x
Formula : Area = Length * breadth
800 = x * 2x
800 = 2x^2
x^2 = 800 / 2
x^2 = 400
x = + _/400
x = + 20
we got x = 20 or x = -20
Since breadth cannot be in negative ,,ignore x = -20 value
Thus, Breadth of mango grove (X) = 20m
Length of mango grove = 2x = 2(20) = 40m
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4) The sum of the ages of two friends is 20yrs. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present age.
sol) Let age of 1st friend be = x
Given :- Sum of the ages = 20yrs
i.e, x + y = 20
=> 2nd friend age (y) = 20 -x
Four years ago
1st friend age = x - 4
2nd friend age = (20 - x) - 4 = 16 -x
Product of their ages 4 yrs ago = 48
(x - 4) ( 16 - x) = 48
x(16 -x) -4(16 -x ) = 48
16x - x^2 - 64 + 4x = 48
-x^2 + 20x - 64 - 48 = 0
-x^2 + 20x - 112 = 0
x^2 - 20x + 112 = 0 { ax^2 + bx + c = 0}
Where a = 1, b = -20, c = 112
We have
D = b^2 - 4ac
D = (-20)^2 - 4* 1* 112
D = 400 - 448
D = -48
Since D < 0
The equation has no real roots
Hence, the given situation is not possible
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5) Is it possible to design a rectangular park of perimeter 80m, and area 400m.sq. If so, find its length and breadth.
sol) Let length be = x
Given :- perimeter = 80m
We know perimeter of rectangle = 2(L + b)
=> 2(X + b) = 80
=> X + b = 80/2
=> b = 40 -x
.^. Breadth be(b) = 40 -x
Given :- Area = 400m.sq
We know area of rectangle = L * B
Area = L * b
400 = x(40 -x)
400 = 40x - x^2
x^2 - 40x + 400 = 0
Factorize :-
x^2 - 20x - 20x + 400 = 0
x(x - 20) -20(x - 20) = 0
(x - 20) ( x - 20) = 0
so, x = 20 is the solution.
.^. Length(X) = 20m
Breadth = 40 - x => 40 -20 = 20m
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(*) Optional Exercise
1) A two digit number is such that the product of the digits is 8. When 18 is added to the number they interchange their places. determine the number.
sol) Let the ten's digit be 'x' and unit's digit b 'y'
.^. Number = 10x +y
Given: product of the digits = 8
xy = 8
y = 8 --------(1)
x
Given: when '18' is added to number they interchange their places.
10x +y+18 = 10y +x
9x - 9y +18=0
x - y + 2= 0-----(2)
put(1) in (2)
x - (8) +2 =0
x
x^2 - 8 + 2x = 0
x^2 +2x -8 = 0
x^2+ 4x-2x-8 = 0
x(x+4)-2(x+4)=0
x = 2 or x=-4
rejecting x=-4 , we have x = 2
put x=2 in (1)
y= 2 =1
2
Thus, the required number is
10x +y = 0
10*2 + 1 = 21 is the required number
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2) A piece of wire 8m. in length, cut into two pieces, and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2m^2?
( Hint : x+y=8, (x/4)^2 +(y/4)^2 = 2
=> (x/4)^2 + ((8-x)/4))^2 = 2 ]
sol) Using Hint we can solve directly
=> x^2 + (64+x^2-16x) =2
16 16
=> x^2 + 64 + x^2 - 16x = 32
=> 2x^2 - 16x +32 = 0
=> x^2 - 8x + 16 = 0
=> x^2 -4x -4x +16 = 0
=> x(x-4) -4(x-4)= 0
=> x = 4
Length of one of the piece of wire = 4m
Given : x +y = 8
4 + y = 8
y = 4
length of another piece of wire=4m
.^.,the cut in the wire should be made exactly middle of the wire.
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3). vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take vinay to complete the job by himself.
sol) praveena can complete the job = x days
Per day praveena can do the job = 1
x
Given:- Vinay can complete the job in 5 days less than praveen
i.e, v=x-5 ---(1)
per day vinay can do the job = 1
(x-5)
Working together they can do the job in 6 days.
It means per day they do the job together
= 1
6
Total work in one day by vinay and praveen
1 + 1 = 1
x (x-5) 6
6[(x-5) + x] = x(x-5)
6[ 2x-5] = x^2 - 5x
12x - 30 = x^2 - 5x
x^2 - 17x +30 = 0
x^2 - 15x -2x + 30 = 0
x(x-15) -2(x -15) = 0
x = 2 or x = 15
But its given that vinay can do the job 5 days faster than praveen
put x = 2 in (1)
v = 2 - 5 = -3 days
put x = 15 in (1)
v = 15 -5 = 10
we take v = 10 as work cannot b done in negative(-3 days)
.^., Vinay can complete work in 10 days
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4) The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2*(16/21), find the fraction.
sol) Given: Denominator of a fraction is one(+1) more than twice(2) the numerator(x).
Let Numerator = x
denominator = one more twice(2x) the numerator= 2x + 1
original Fraction = x
2x + 1
mixed fraction = 2(16/21) = 42+16 = 58
21 21
Given: sum of fraction and its reciprocal = 58
21
so, the following balanced equation will be formed.
fraction + reciprocal
x + 2x +1 = 58
2x+1 x 21
x^2 + 4x^2 + 4x + 1 = 58
x(2x+1) 21
21(x^2 + 4x^2 + 4x +1) = 58(2x^2 + x)
21(5x^2 + 4x + 1) = 116x^2 + 58x
105x^2 + 84x + 21 = 116x^2 + 58x
11x^2 - 26x - 21 = 0
11x^2 - 33x +7x - 21 = 0
11x(x - 3) + 7(x -3) = 0
(11x+7) = 0 (x-3) = 0
x = -7/11 or x = 3
we will take x = 3
Now
denominator = 2x +1 = 2(3) +1
=> 6 +1 = 7
Thus, fraction is = 3/7
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