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RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations Ex 8.3 Download PDF

Chapter 8: Quadratic Equations Exercise – 8.3

Question: 1

Find the roots of the equation (x – 4) (x + 2) = 0

Solution:

The given equation is (x – 4) (x + 2) = 0

Either x – 4 = 0 therefore x = 4

Or, x + 2 = 0, therefore x = – 2

The roots of the above mentioned quadratic equation are 4 and -2 respectively. 

Question: 2

Find the roots of the equation (2x + 3) (3x – 7) = 0

Solution:

The given equation is (2x + 3) (3x – 7) = 0.

Either 2x + 3 = 0,

Therefore x = – 3/2  

Or, 3x -7 = 0, therefore x = 7/3 

The roots of the above mentioned quadratic equation are x = -3/2 and x = 7/3 respectively. 

Question: 3

Find the roots of the quadratic equation 3x² – 14x – 5 = 0

Solution:

The given equation is 3x² – 14x – 5 = 0

= 3x² – 14x – 5 = 0

= 3x² – 15x + x – 5 = 0

= 3x(x – 5) + 1(x – 5) = 0 = (3x + 1)(x – 5) = 0

Either 3x + 1 = 0 therefore x = -1/3  

Or, x-5 =0 therefore x = 5

The roots of the given quadratic equation are 5 and x = – 1/3 respectively. 

Question: 4

Find the roots of the equation 9x² – 3x – 2 = 0.

Solution:

The given equation is 9x² – 3x – 2 = 0.

= 9x² – 3x – 2 = 0.

= 9x² – 6x + 3x – 2 = 0

= 3x (3x – 2) + 1(3x – 2) = 0

= (3x – 2)(3x + 1) = 0

Either, 3x – 2 = 0 therefore x = 2/3

Or, 3x + 1= 0  therefore x = -1/3 

The roots of the above mentioned quadratic equation are x = 2/3 and x = -1/3 respectively.   

Question: 5

Find the roots of the quadratic equation

Solution:

The given equation is

Cancelling out the like terms on both the sides of the numerator. We get,

= x²+ 4x – 5 = 7

= x²+ 4x – 12 = 0

= x² + 6x – 2x – 12 = 0

= x(x + 6) – 2(x – 6) = 0

= (x + 6)(x – 2) = 0

Either x + 6 = 0

Therefore x = -6 Or, x – 2 = 0

Therefore x = 2

The roots of the above mentioned quadratic equation are 2 and – 6 respectively

Question: 6

Find the roots of the equation 6x² + 11x + 3 = 0.

Solution:

The given equation is 6x² + 11x + 3 = 0.

= 6x² + 11x + 3 = 0.

= 6x² + 9x + 2x + 3 = 0

= 3x (2x + 3) + 1(2x + 3) = 0

= (2x +3) (3x + 1) = 0

Either, 2x + 3 = 0

Therefore x = -3/2

 Or, 3x + 1= 0 therefore x = -1/3 

The roots of the above mentioned quadratic equation are x = -3/2 and x = -1 /3 respectively. 

Question: 7

Find the roots of the equation 5x² – 3x – 2 = 0

Solution:

The given equation is 5x² – 3x – 2 = 0.

= 5x² – 3x – 2 = 0.

= 5x² – 5x + 2x – 2 = 0

= 5x(x – 1) + 2(x – 1) = 0

= (5x + 2)(x – 1) = 0

Either 5x + 2 = 0 therefore 

x = -2/5 Or, x -1= 0

Therefore x = 1

The roots of the above mentioned quadratic equation are 1 and x = -2/5 respectively. 

Question: 8

Find the roots of the equation 48x² – 13x – 1 = 0

Solution:

The given equation is 48x² – 13x – 1 = 0.

= 48x² – 13x – 1 = 0.

= 48x² – 16x + 3x – 1 = 0.

= 16x(3x – 1) + 1(3x – 1) = 0

= (16x + 1)(3x – 1) = 0

Either 16x + 1 = 0 therefore x = -1/16  

Or, 3x-1=0 Therefore x = 1/3 

The roots of the above mentioned quadratic equation are x = -1/16 and x = 1/3 respectively. 

Question: 9

Find the roots of the equation 3x² = -11x – 10 

Solution:

The given equation is 3x² = -11x – 10 = 3x²

= -11x – 10 = 3x² + 11x + 10 = 0

= 3x² + 6x + 5x + 10 = 0

= 3x(x + 2) + 5(x + 2) = 0

= (3x + 2)(x + 2) = 0

Either 3x + 2 = 0 therefore x = -2/3

Or, x + 2 = 0 therefore x = -2

The roots of the above mentioned quadratic equation are x = -2/3 and -2 respectively. 

Question: 10

Find the roots of the equation 25x(x + 1) = – 4 

Solution:

The given equation is 25x(x + 1) = 4

= 25x(x + 1) = -4

= 25x² + 25x + 4 = 0

= 25x² + 20x + 5x + 4 = 0

= 5x (5x + 4) + 1(5x + 4) = 0

= (5x + 4)(5x + 1) = 0

Either 5x + 4 = 0 therefore x = – 4/5

Or, 5x + 1 = 0 therefore x = -1 /5

The roots of the quadratic equation are x = – 4/5 and x = (-1)/5 respectively. 

Question: 11

Find the roots of the quadratic equation

Solution:

The given equation is

Cross multiplying both the sides.

We get,

= 2 = 3x(x – 2)

= 2 = 3x² – 6x

= 3x²– 6x – 2 = 0

= 3x²– 3x – 3x – 2 = 0

The roots of the above mentioned quadratic equation are 

Question: 12

Find the roots of the quadratic equation

Solution:

The given equation is

 The roots of the above mentioned quadratic equation are

Question: 13

Find the roots of the quadratic equation

Solution:

The given equation is

Cancelling out the like numbers on both the sides of the equation

= x² – 3x – 28 = – 30

= x² – 3x – 2 = 0

= x² – 2x – x – 2 = 0

= x(x – 2) – 1(x – 2) = 0

= (x – 2)(x – 1) = 0

Either x – 2 = 0

Therefore x = 2 Or, x-1 = 0

Therefore x = 1

The roots of the above mentioned quadratic equation are 1 and 2 respectively. 

Question: 14

Find the roots of the quadratic equation a²x² – 3abx + 2b² = 0

Solution:

The given equation is a²x² – 3abx + 2b² = 0

= a²x² – 3abx + 2b² = 0

= a²x² – abx – 2abx + 2b² = 0

= ax(ax – b) – 2b(ax – b) = 0

= (ax – b)(ax – 2b) = 0

Either ax – b = 0

Therefore x = b/a Or, ax – 2b = 0

Therefore x = 2/ba   

The roots of the quadratic equation are x = 2/ba and x = b/a respectively. 

Question: 15

Find the roots of the 4x² + 4bx – (a² – b²) = 0 

Solution:

= 4(a² – b²) = -4(a – b)(a + b) = -2(a-b)*2(a + b)

=2(b – a)*2(b + a) = 4x²+ (2(b – a) + 2(b + a)) – (a – b)(a + b) = 0

= 4x²  + 2(b – a)x++ 2(b + a)x + (b – a)(a + b) = 0

= 2x(2x + (b – a)) +(a + b)(2x + (b – a)) = 0

= (2x + (b – a))(2x + b + a) = 0

Either, (2x + (b – a)) = 0

Therefore x = (a – b)/2 Or, (2x + b + a) = 0

Therefore x = (- a – b)/2   

The roots of the above mentioned quadratic equation are x = (- a – b)/2 and x = (a – b)/2 respectively.

Question: 16

Find the roots of the equation ax² + (4a² – 3b)x – 12ab = 0

Solution:

The given equation is ax² + (4a² – 3b)x – 12ab = 0

= ax² + (4a² – 3b)x – 12ab = 0

= ax² + 4a²x – 3bx – 12ab = 0

= ax(x – 4a) – 3b(x – 4a) = 0

= (x – 4a)(ax – 4b) = 0

Either x – 4a = 0

Therefore x = 4/a Or, ax – 4b = 0

Therefore x = 4b/a 

The roots of the above mentioned quadratic equation are x = 4b/a and 4a respectively.

Question: 17

Find the roots of

Solution:

The given equation is

= (x + 3)(2x – 3) = (x + 2)(3x – 7)

= 2x² – 3x + 6x – 9 = 3x² – x – 14

= 2x² + 3x – 9 = 3x² – x – 14

= x² – 3x – x – 14 + 9 = 0

= x² – 5x + x – 5 = 0

= x(x – 5) + 1(x – 5) = 0

= (x – 5)(x + l) – 0

Either x – 5 – 0 or x + 1 = 0 x = 5 and x = -1

The roots of the above mentioned quadratic equation are 5 and -1 respectively. 

Question: 18

Find the roots of the equation

Solution:

The given equation is

= 3(4x² – 19x + 20) = 25(x² – 7x + 12)

= 12x² – 57x + 60 = 25x² – 175x + 300

= 13x² – 78x – 40x + 240 = 0

= 13x² – 118x + 240 = 0

= 13x² – 78x – 40x + 240 = 0

= 13x(x – 6) – 40(x – 6) = 0

= (x – 6)(13x – 40) = 0

Either x – 6 = 0

Therefore x = 6 Or, 13x – 40 = 0

Therefore x = 40/13 

The roots of the above mentioned quadratic equation are 6 and 40/13 respectively. 

Question: 19

Find the roots of the quadratic equation

Solution:

The given equation is

= 4(2x² + 2) = 17(x² – 2x)

= 8x² + 8 = 17x² – 34x

= 9x² – 34x – 8 = 0

= 9x² – 36x + 2x – 8 = 0

= 9x(x – 4) + 2(x – 4) = 0

= (9x + 2)(x – 4) = 0

Either 9x + 2 = 0

Therefore x = (-2)/9 Or, x – 4 = 0

Therefore x = 4

The roots of the above mentioned quadratic equation are x = (- 2)/9 and 4 respectively. 

Question: 20

Find the roots of the quadratic equation

Solution:

The equation is

= x(3x – 5) = 6(x² – 3x + 2)

= 3x² – 5x = 6x² – 18x + 12

= 3x² – 13x + 12 = 0

= 3x² – 9x – 4x + 12 = 0

= 3x(x – 3) – 4(x – 3) = 0

= (x – 3)(3x – 4) = 0

Either x – 3 = 0

Therefore x = 3 Or, 3x – 4 = 0

Therefore 4/3. The roots of the above mentioned quadratic equation are 3 and 4/3 respectively.

Question: 21

Find the roots of the quadratic equation

Solution:

The equation is

= 6(4x) = 5(x² – 1) = 24x

= 5x² – 5 = 5x² – 24x – 5 =0

= 5x² – 25x + x – 5 = 0

= 5x(x – 5) + 1(x – 5) = 0

= (5x + 1)(x – 5) = 0

Either x – 5 = 0

Therefore x = 5 Or, 5x + 1 = 0

Therefore x = −1/5 

The roots of the above mentioned quadratic equation are x = −1/5 and 5 respectively. 

Question: 22

Find the roots of the quadratic equation

Solution:

The equation is

= 2(5x² + 2x + 2) = 5(2x² – x – 1)

= 10x² + 4x + 4 = 10x² – 5x – 5

Cancelling out the equal terms on both sides of the equation.

We get, = 4x + 5x + 4 + 5 = 0

= 9x + 9 = 0

= 9x = -9 X = -1 X = -1 is the only root of the given equation.

Question: 23

Find the roots of the quadratic equation

Solution:

The given equation is

= m²x² + 2mnx + (n² – mn) = 0

Now we solve the above quadratic equation using factorization method

Therefore

Now, one of the products must be equal to zero for the whole product to be zero for the whole product to be zero. Hence, we equate both the products to zero in order to find the value of x.

The roots of the above mentioned quadratic equation are

Question: 24

Find the roots of the quadratic equation

Solution:

The given equation is

= (2x² – 2x(a + b) + a² + b²)ab

= (a² + b²)(x² – (a + b)x + ab)

= (2abx² – 2abx(a + b) + ab(a² + b²))

= (a² + b²)(x² – (a² + b²)(a + b)x + (a² + b²)(ab)

= (a² + b² – 2ab)x – (a + b)(a² + b² – 2ab)x = 0

= (a – b)²x² – (a + b)(a + b)²x² = 0

= x(a – b)²(x – (a + b)) = 0

= x(x – (a + b)) = 0

Either x = 0 Or, (x – (a + b)) = 0

Therefore x = a + b

The roots of the above mentioned quadratic equation are 0 and a + b respectively.

Question: 25

Find the roots of the quadratic equation

Solution:

The given equation is

Cancelling out the like terms on both the sides of numerator and denominator.

We get,

= (x – 1)(x – 4) = 18

= x² – 4x – x + 4 = 18

= x² – 5x – 14 = 0

= x² – 7x + 2x – 14 = 0

= x(x – 7) + 2(x – 7) = 0

= (x – 7)(x + 2) = 0

Either x – 7 = 0

Therefore x = 7 Or, x + 2 = 0

Therefore x = -2

The roots of the above mentioned quadratic equation are 7 and -2 respectively. 

Question: 26

Find the roots of the quadratic equation

Solution:

The given equation is

= (x – c)(ax – 2ab + bx)

= 2c(x² – bx – ax + ab)

= (a + b)x² – 2abx – (a + b)cx + 2abc

= 2cx² – 2c(a + b)x + 2abc 

Question: 27

Find the roots of the Question x² + 2ab = (2a + b)x

Solution:

The given equation is x² + 2ab = (2a + b)x

= x² + 2ab = (2a+b)x

= x² – (2a + b)x + 2ab = 0

= x² – 2ax – bx + 2ab = 0

= x(x – 2a) – b(x – 2a) = 0

= (x – 2a)(x – b) = 0

Either x – 2a = 0

Therefore x = 2a Or, x – b = 0

Therefore x = b

The roots of the above mentioned quadratic equation are 2a and b respectively.

Question: 28

Find the roots of the quadratic equation (a + b)²x² – 4abx – (a – b)²  = 0

Solution:

The given equation is (a + b)²x² – 4abx – (a – b)²  = 0

= (a + b)²x² – 4abx – (a – b)²  = 0  

= (a + b)²x² – ((a + b)² – (a – b)²)x – (a – b)² = 0

= (a + b)²x² – (a + b)² x + (a – b)² x – (a – b)² = 0

= (a + b)²x(x – 1) (a + b)² (x – 1) = 0

= (x – 1) (a + b)²x +( a + b)²) = 0

Either x – 1 = 0

Therefore x = 1 Or, (a + b)²x + (a + b)²) = 0

The roots of the above mentioned quadratic equation are –and 1 respectively. 

Question: 29

Find the roots of the quadratic equation a(x² + 1) – x(a² + 1) = 0

Solution:

The given equation is a(x² + 1) – x(a² + 1) = 0

= ax² + a – a²x – x = 0

= ax(x – a) – 1(x – a) = 0

= (x – a)(ax – 1) = 0

Either x – a = 0

Therefore x = a Or, ax – 1 = 0

Therefore x = 1/a 

The roots of the above mentioned quadratic equation are  (a) and x = 1/a respectively. 

Question: 30

Find the roots of the quadratic equation

Solution:

The given equation is 

Either x + a = 0

Therefore x = -a Or, (x + 1/a) = 0 

Therefore x = 1/a. The roots of the above mentioned quadratic equation are x = 1/a and – a respectively. 

Question: 31

Find the roots of the quadratic equation abx² + (b² – ac)x – bc = 0 

Solution:

The given equation is abx² + (b² – ac)x – bc = 0

= abx² + b²x – acx – bc = 0

= bx (ax + b) – c (ax + b) = 0

= (ax + b)(bx – c) = 0

Either, ax + b = 0

Therefore x = −b/a Or, bx – c = 0

Therefore x = c/b

The roots of the above mentioned quadratic equation are x = c/b and x = −b/a respectively. 

Question: 32

Find the roots of the quadratic equation a²b²x² + b²x – a²x – 1 = 0

Solution:

The given equation is a²b²x² + b²x – a²x – 1 = 0

= b²x(a²x + 1) -1 (a²x + 1) = (a²x+1)( b²x-1) =0

Either (a²x + 1) = 0

Therefore x = -1/a²Or, ( b²x-1) = 0

Therefore x = -1/b²

The roots of the above mentioned quadratic equation are x = 1/b² and x = (-1)/a²respectively. 

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