RD Sharma Solutions Chapter 8 Quadratic Equations Exercise 8.3 Class 10 Maths

Chapter Name	RD Sharma Chapter 8 Quadratic Equations
Book Name	RD Sharma Mathematics for Class 10
Other Exercises	Exercise 8.1 Exercise 8.2 Exercise 8.4 Exercise 8.5 Exercise 8.6
Related Study	NCERT Solutions for Class 10 Maths

Exercise 8.3 Solutions

Solve the following quadratic equations by factorization:

1. (x - 4)(x + 2) = 0

Solution

We have
(x - 4)(x + 2) = 0
⇒ either (x - 4) = 0 or (x + 2) = 0
⇒ x = 4 or x = -2
Thus, x = 4 and x = -2 are two roots of the equation(x - 4)(x + 2) = 0

2. (2x + 3)(3x - 7) = 0

Solution

We have,
(2x + 3)(3x- 7) = 0
⇒ (2x + 3) = 0 or (3x- 7) = 0
⇒ 2x = - 3 or 3x = 7
⇒ x = -3/2 or x = 7/3
Thus, x = -3/2 and x = 7/3 are two roots of the equation(2x + 3)(3x - 7) = 0

3. 4x² + 5x = 0

Solution

We have 4x² + 5x = 0
⇒ x (4x + 5) = 0
⇒ either x = 0 or 4x + 5 = 0
⇒ x = 0 or 4x = -5
⇒ x = 0 or x = -5/4
Thus, x = 0 and x = -5/4 are two roots of equation 4x² + 5x = 0

4. 9x² - 3x - 2 = 0

Solution

We have 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 or 3x + 1 = 0
⇒ either 3x - 2 = 0 or 3x + 1 = 0
⇒ 3x = 2 or 3x = - 1
⇒ x = 2/3 or x = -1/3
Thus, x = 2/3 and x = -1/3 are two roots of the equation 9x² - 3x - 2 = 0

5. 6x² - x - 2 = 0

Solution

We have 6x² - x - 2 = 0
⇒ 6x² + 3x - 4x - 2 = 0
⇒ 3x(2x + 1) - 2(2x + 1) = 0
⇒ (2x + 1)(3x - 2) = 0
⇒ either 2x + 1 = 0 or 3x - 2 = 0
⇒ 2x = - 1 or 3x = 2
⇒ x = -1/2 or x = 2/3
Thus, x = -1/2 and x = 2/3 are two roots of the equation 6x² - x - 2 = 0

6. 6x² + 11x + 3 = 0

Solution

We have
6x² + 11x + 3 = 0
⇒ 6x² + 9x + 2x + 3 = 0
⇒ 3x(2x + 3) + 1 (2x + 3) = 0
⇒ (2x + 3)(3x + 1) = 0
⇒ 2x + 3 = 0 or x = -1/3
Thus, x = -3/2 and x = -1/3 are the two roots of the given equation.

7. 5x² - 3x - 2 = 0

Solution

We have
5x² - 3x - 2 = 0
⇒ 5x² - 5x + 2x - 2 = 0
⇒ 5x(x - 1) + 2(x - 1) = 0
⇒ (x - 1)(5x + 2) = 0
⇒ (x - 1) = 0 or 5x + 2 = 0
⇒ x = 1 or x = -2/5
∴ x = 1 and x = -2/5 are the two roots of the given equation.

8. 48x² - 13x - 1 = 0

Solution

We have
48x² - 13x - 1 = 0
⇒ 48x² - 16x + 3x - 1 = 0
⇒ 16x (3x - 1) + 1(3x - 1) = 0
⇒ (3x - 1)(16x + 1) = 0
⇒ 3x - 1 = 0 or 16x + 1 = 0
⇒ x = 1/3 or x = -1/16
∴ x= -1/16 and x = 1/3 are the two roots of the given equation.

9. 3x² = - 11x - 10

Solution

We have
3x² = - 11x - 10
⇒ 3x² + 11x + 10 = 0
⇒ 3x² + 6x + 5x + 10 = 0
⇒ 3x(x + 2) + 5(x + 2) = 0
⇒ (x + 2)(3x + 5) = 0
⇒ (x + 2) = 0 or x = -5/3
∴ x = -2 and x = -5/3 are the two roots at the quadratic equation 3x² = -11x - 10

10. 25x(x + 1) = - 4

Solution

We have
(x + 1) = -4
⇒(25x) × x + (25x) × 10 - 4
⇒ 25x² + 25x + 4 = 0 [25×4 = 100 ⇒ 25 = 20 + 5 ⇒ 100 = 20×5]
⇒ 25x² + 20x + 5x + 4 = 0
⇒ 5x(5x + 4) + 1(5x + 4) = 0
⇒ (5x + 4)(5x + 1) = 0
⇒ 5x + 4 = 0 or 5x + 1 = 0
⇒ x = -4/3 or x = -1/3
∴ x = -4/3 and x = -1/5 are the two solutions of the quadratic equation 25x(x + 1) = -4

11. 10x - 1/x = 3

Solution

We have

12. 2/2² - 5/x + 2 = 0

Solution

13. 4√3x² + 5x - 2√3 = 0

Solution

We have,
4√3x² + 5x - 2√3 = 0

14. √2x² - 3x - 2√2 = 0

Solution

We have,

15. a²x² - 30bx + 2b² = 0

Solution

We have,

16. x² - (√2 + 1)x + √2 = 0

Solution

We have,

∴ x = 1 and x = √2 are the roots of the given quadratic equation

17. x² - (√3 + 1)x + √3 = 0

Solution

x² - (√3 + 1)x + √3 = 0

18. 4x² + 4bx - (a² - b²) = 0

Solution

19. ax² + (4a² - 3b)x - 12ab = 0

Solution

We have,

ax² + (4a² - 3b)x - 12ab = 0

20. (x - 1/2)² = 4

Solution

We have,

21. x² - 4√2 x + 6 = 0

Solution

We have,

22. (x + 3)/(x + 2) = (3x - 7)/(2x - 3)

Solution

⇒ (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 - 4x - 5 = 0
[1x - 5 = -5 - 4 = -5 + 1]
⇒ x² - 5x + x - 5 = 0
⇒ x(x - 5) + 1(x - 5) = 0
⇒ (x - 5)(x + 1) = 0
⇒ x - 5 = 0 or x + 1 = 0
⇒ x = 5 or x = - 1
∴ x = 5 and x = - 1 are the two roots of the given quadratic equation.

23 2x/(x - 4) + (2x -5)/(x- 3) = 25/3

Solution