RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.3 (Updated for 2024)

Solve the following quadratic equation by factorization:

1. (x – 4)(x + 2) = 0

Solution:

Given,

(x – 4) (x + 2) = 0

So, either x – 4 = 0 ⇒ x = 4

Or, x + 2 = 0, ⇒ x = – 2

Thus, the roots of the given quadratic equation are 4 and -2, respectively.

2. (2x + 3) (3x – 7) = 0

Solution:

Given,

(2x + 3) (3x – 7) = 0.

So, either 2x + 3 = 0, ⇒ x = – 3/2

Or, 3x -7 = 0, ⇒ x = 7/3

Thus, the roots of the given quadratic equation are x = -3/2 and x = 7/3, respectively.

3. 3x² – 14x – 5 = 0

Solution:

Given.

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

Now, either 3x + 1 = 0 ⇒ x = -1/3

Or, x – 5 = 0 ⇒ x = 5

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

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

Solution:

Given,

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

Now, either 3x – 2 = 0 ⇒ x = 2/3

Or, 3x + 1= 0 ⇒ x = -1/3

Thus, the roots of the given quadratic equation are x = 2/3 and x = -1/3, respectively.

5.

Solution:

Given,

Dividing by 6 on both sides and cross-multiplying, we get

x²+ 4x – 12 = 0

⇒ x² + 6x – 2x – 12 = 0

⇒ x(x + 6) – 2(x – 6) = 0

⇒ (x + 6)(x – 2) = 0

Now, either x + 6 = 0 ⇒x = -6

Or, x – 2 = 0 ⇒ x = 2

Thus, the roots of the given quadratic equation are 2 and – 6, respectively.

6. 6x² + 11x + 3 = 0

Solution:

Given equation is 6x² + 11x + 3 = 0.

⇒ 6x² + 9x + 2x + 3 = 0

⇒ 3x (2x + 3) + 1(2x + 3) = 0

⇒ (2x +3) (3x + 1) = 0

Now, either 2x + 3 = 0 ⇒ x = -3/2

Or, 3x + 1= 0 ⇒ x = -1/3

Thus, the roots of the given quadratic equation are x = -3/2 and x = -1/3, respectively.

7. 5x² – 3x – 2 = 0

Solution:

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

Now, either 5x + 2 = 0 ⇒x = -2/5

Or, x -1= 0 ⇒x = 1

Thus, the roots of the given quadratic equation are 1 and x = -2/5, respectively.

8. 48x² – 13x – 1 = 0

Solution:

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 ⇒ x = -1/16

Or, 3x – 1=0 ⇒ x = 1/3

Thus, the roots of the given quadratic equation are x = -1/16 and x = 1/3, respectively.

9. 3x² = -11x – 10 

Solution:

Given equation is 3x² = -11x – 10

⇒ 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 5)(x + 2) = 0

Now, either 3x + 5 = 0 ⇒ x = -5/3

Or, x + 2 = 0 ⇒ x = -2

Thus, the roots of the given quadratic equation are x = -5/3 and -2, respectively.

10. 25x(x + 1) = – 4 

Solution:

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

Now, either 5x + 4 = 0 therefore x = – 4/5

Or, 5x + 1 = 0 therefore x = -1 /5

Thus, the roots of the given quadratic equation are x = – 4/5 and x = -1/5, respectively.

11. 16x – 10/x = 27

Solution:

Given,

16x – 10/x = 27

On multiplying x on both sides we have,

⇒ 16x² – 10 = 27x

⇒ 16x² – 27x – 10 = 0

⇒ 16x² – 32x + 5x – 10 = 0

⇒ 16x(x – 2) +5(x – 2) = 0

⇒ (16x + 5) (x – 2) = 0

Now, either 16x + 5 = 0 ⇒ x = -5/16

Or, x – 2 = 0 ⇒ x = 2

Thus, the roots of the given quadratic equation are x = – 5/16 and x = 2, respectively.

12.

Solution:

Given equation is,

On cross multiplying on both the sides we get,

2 = 3x(x – 2)

2 = 3x² – 6x

3x²– 6x – 2 = 0

⇒ 3x²– 3x – 3x – 2 = 0

Now, either

Thus,

are the solutions of the given quadratic equations.

13. x – 1/x = 3, x ≠ 0

Solution:

Given,

x – 1/x = 3

On multiplying x on both sides, we have,

⇒ x² – 1 = 3x

⇒ x² – 3x – 1 = 0

14.

Solution:

Given,

Dividing by 11 both the sides and cross-multiplying, we get,

⇒ 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

Now, either x – 2 = 0 ⇒ x = 2

Or, x – 1 = 0 ⇒ x = 1

Thus, the roots of the given quadratic equation are 1 and 2, respectively.

15.

Solution:

Given,

On cross multiplying we get,

⇒ x(3x – 8) = 8(x – 3)(x – 2)

⇒ 3x² – 8x = 8(x² – 5x + 6)

⇒ 8x² – 40x + 48 – (3x² – 8x) = 0

⇒ 5x² – 32x + 48 = 0

⇒ 5x² – 20x – 12x + 48 = 0

⇒ 5x(x – 4) – 12(x – 4) = 0

⇒ (x – 4)(5x – 12) = 0

Now, either x – 4 = 0 ⇒ x = 4

Or, 5x – 12 = 0 ⇒ x = 12/5

Thus, the roots of the given quadratic equation are 12/5 and 4, respectively.

16. a²x² – 3abx + 2b² = 0

Solution:

Given equation is a²x² – 3abx + 2b² = 0

⇒ a²x² – abx – 2abx + 2b² = 0

⇒ ax(ax – b) – 2b(ax – b) = 0

⇒ (ax – b)(ax – 2b) = 0

Now, either ax – b = 0 ⇒ x = b/a

Or, ax – 2b = 0 ⇒ x = 2b/a

Thus, the roots of the quadratic equation are x = 2b/a and x = b/a, respectively.

17. 9x² – 6b²x – (a⁴ – b⁴) = 0

Solution:

Given,

Thus, the roots of the quadratic equation are x = (b² – a²)/3 and x = (a² + b²)/3, respectively.

18. 4x² + 4bx – (a² – b²) = 0 

Solution:

Given,

4x² + 4bx – (a² – b²) = 0

For factorizing,

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

So, now

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

Now, either (2x + (b – a)) = 0 ⇒x = (a – b)/2

Or, (2x + b + a) = 0 ⇒ x = -(a + b)/2

Thus, the roots of the given quadratic equation are x = -(a + b)/2 and x = (a – b)/2, respectively.

19. ax² + (4a² – 3b)x – 12ab = 0

Solution:

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

⇒ ax² + 4a²x – 3bx – 12ab = 0

⇒ ax(x + 4a) – 3b(x + 4a) = 0

⇒ (x + 4a)(ax – 3b) = 0

Now, either x + 4a = 0 ⇒ x = -4a

Or, ax – 3b = 0 ⇒ x = 3b/a

Thus, the roots of the given quadratic equation are x = 3b/a and -4a, respectively.

20. 2x² + ax – a² = 0

Solution:

Given,

Thus, the roots of the given quadratic equation are x = a/2 and -a, respectively.

21. 16/x – 1 = 15/(x + 1), x ≠ 0, -1

Solution:

Given,

Thus, the roots of the given quadratic equation are x = 4 and -4, respectively.

22. , x ≠ -2, 3/2

Solution:

Given,

On cross-multiplying we get,

(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

Now, either x – 5 = 0 or x + 1 = 0

⇒ x = 5 and x = -1

Thus, the roots of the given quadratic equation are 5 and -1, respectively.

23.

, x ≠ 3, 4

Solution:

The given equation is

On cross-multiplying, we have

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

Now, either x – 6 = 0 ⇒x = 6

Or, 13x – 40 = 0 ⇒x = 40/13

Thus, the roots of the given quadratic equation are 6 and 40/13, respectively.

24. x ≠ 0, 2

Solution:

Given equation is,

On cross-multiplying, we get

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

Now, either 9x + 2 = 0 ⇒x = -2/9

Or, x – 4 = 0 ⇒ x = 4

Thus, the roots of the given quadratic equation are x = -2/9 and 4, respectively.

25.

Solution:

Given equation is,

On cross-multiplying, we get

7(-12x) = 48(x² – 9)

⇒ -84x = 48x² – 432

⇒ 48x² + 84x – 432 = 0

⇒ 4x² + 7x – 36 = 0 dividing by 12]

⇒ 4x² + 16x – 9x – 36 = 0

⇒ 4x(x + 4) – 9(x – 4) = 0

⇒ (4x – 9)(x + 4) = 0

Now, either 4x – 9 = 0 ⇒x = 9/4

Or, x + 4 = 0 ⇒ x = -4

Thus, the roots of the given quadratic equation are x = 9/4 and -4, respectively.

26.

, x ≠ 0

Solution:

Given equation is,

On cross multiplying, we have

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

Now, either x – 3 = 0 ⇒ x = 3

Or, 3x – 4 = 0 ⇒ 4/3.

Thus, the roots of the given quadratic equation are 3 and 4/3, respectively.

27.

, x ≠ 1, -1

Solution:

The given equation is,

On cross – multiplying we have,

⇒ 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

Now, either x – 5 = 0 ⇒ x = 5

Or, 5x + 1 = 0 ⇒ x = −1/5

Thus, the roots of the given quadratic equation are x = −1/5 and 5, respectively.

28.

, x ≠ 1, -1/2

Solution:

The given equation is,

On cross – multiplying we have,

⇒ 2(5x² + 2x + 2) = 5(2x² – x – 1)

⇒ 10x² + 4x + 4 = 10x² – 5x – 5

⇒ 4x + 5x + 4 + 5 = 0

⇒ 9x + 9 = 0

⇒ 9x = -9

Thus, x = -1 is the only root of the given equation.

29.

Solution:

Given equation is,

Thus, the roots of the given quadratic equation are x = 1 and x = -2, respectively.

30.

Solution:

Given equation is,

On cross-multiplying, we have,

3(2x² – 22x + 58) = 10(x² – 12x + 35)

⇒ 6x² – 66x + 174 = 10x² – 120x + 350

⇒ 4x² – 54x + 176 = 0

⇒ 2x² – 27x + 88 = 0

⇒ 2x² – 16x – 11x + 88 = 0

⇒ 2x(x – 8) – 11(x + 8) = 0

⇒ (x – 8)(2x – 11) = 0

Now, either x – 8 = 0 ⇒ x = 8

Or, 2x – 11 = 0 ⇒ x = 11/2

Thus, the roots of the given quadratic equation are x = 11/2 and 8, respectively.