Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1

Solve the following quadratic equations by factorization

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

Solution:

We have equation, 

(x - 4) (x + 2) = 0  

Implies that either x - 4 = 0 or x + 2 = 0

Therefore, roots of the equation are 4 or -2.

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

Solution:

We have equation,

(2x + 3) (3x - 7) = 0  

Implies that either 2x + 3 = 0 or 3x - 7 = 0

Therefore, roots of the equation are -3/2 or 7/3.

Question 3. 3x² - 14x - 5 = 0

Solution:

We have equation,

3x² - 14x - 5 = 0

We can factorize this equation as 

3x² - 15x + x - 5 = 0  

3x (x - 5) -1 (x - 5) = 0

(3x - 1) (x - 5) = 0 

Therefore, roots of the equation are -1/3 or 5.

Question 4. 9x² - 3x - 2 = 0

Solution:

We have equation,

9x² - 3x - 2 = 0

We can factorize this equation as

9x² - 6x + 3x - 2 = 0  

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

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

Therefore, roots of the equation are 2/3 or -1/3.

Question 5. (1/(x - 1)) - (1/(x + 5)) = 6/7, x ≠ 1, -5.

Solution:

We have equation,

 (1/(x - 1)) - (1/(x + 5)) = 6/7 

We can rewrite this equation as

(x + 5 - x + 1)/((x - 1) (x + 5)) = 6 / 7

1/((x - 1) (x + 5)) = 1/7, or

(x - 1) ( x + 5) = 7

x² + 4x - 12 = 0

We can factorize this equation as

x² + 6x - 2x - 12 = 0

x (x + 6) - 2 (x + 6) = 0

(x - 2) (x + 6) = 0

Therefore, roots of the equation are 2 or -6.

Question 6. 6x² + 11x + 3 = 0

Solution:

We have equation,

6x² + 11x + 3 = 0

We can factorize this equation as

6x² + 9x + 2x + 3 = 0  

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

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

Therefore, roots of the equation are -3/2 or -1/3.

Question 7. 5x² - 3x - 2 = 0

Solution:

We have equation,

5x² - 3x - 2 = 0

We can factorize this equation as

5x² - 5x + 2x - 2 = 0  

5x (x - 1) + 2(x - 1) = 0

(x - 1) (5x + 2) = 0

Therefore, roots of the equation are -2/5 or -1.

Question 8. 48x² - 13x - 1 = 0.

Solution:

We have equation,

48x² - 13x - 1 = 0

We can factorize this equation as

48x² - 16x + 3x - 1 = 0  

16x (3x - 1) + 1 (3x - 1) = 0

(3x - 1) (16x + 1) = 0

Therefore, roots of the equation are -1/16 or 1/3.

Question 9. 3x² = -11x - 10.

Solution:

We have equation,

3x² + 11x +10 = 0

We can factorize this equation as

3x² + 6x + 5x + 10 = 0  

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

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

Therefore, roots of the equation are -2 or -5 / 3.

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

Solution:

We have equation,

25x² + 25x + 4 = 0

We can factorize this equation as

25x² + 20x + 5x + 4 = 0  

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

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

Therefore, roots of the equation are -1/5 or -4/5.

Question 11. 16x - (10/x) = 27

Solution:

We have equation 16x² -27x -10 = 0

We can factorize this equation as:

16x² - 32x + 5x - 10 = 0

16x (x - 2) + 5 (x - 2) = 0

(16x + 5) (x - 2) = 0

Therefore, the roots of equations are 2 or -5/16.

Question 12. (1/x) - (1/(x - 2)) = 3, x ≠ 0, 2

Solution:

We have equation 3x² - 6x + 2 = 0

Here, a = 3, b = -6 and c = 2

Since,

Discriminant (D) = b² - 4ac and x = (-b ± √D)/2a

Therefore, 

D = 36 - 24 = 12, and

x  = (-(-6) ± √12)/6

x = (6 ± 2√3)/6

x = (3 ± √3)/3

Therefore, the roots of equations are (3 + √3)/3 or (3 - √3)/3.

Question 13. x - (1/x) = 3, x ≠ 0

Solution:

 We have equation x² - 3x  -1 = 0

Here, a = 1, b = -3 and c = -1

Since,

Discriminant (D) = b² - 4ac and x = (-b ± √D)/2a

Therefore,

D = 9 + 4 = 13, and

x = (-(-3) ± √13)/2

x = (3 ± √13)/2

Therefore, the roots of equations are (3 + √13)/2 or (3 - √13)/2.

Question 14. (1/(x + 4)) - (1/(x - 7)) = 11 / 30, x ≠ 4, 7

Solution:

 We have equation,

(1/(x + 4)) - (1/(x - 7)) = 11/30

(x - 7 - x - 4)/((x + 4) (x - 7)) = 11/30

-11/((x - 4) (x - 7)) = 11/30

-1/((x - 4) (x - 7)) = 1/30

 x² - 3x + 2 = 0

We can factorize this equation as:

x² - 2x - x + 2 = 0

x (x - 2) - 1 (x - 2) = 0

(x -1) (x - 2) = 0

Therefore, the roots of equations are 2 or 1.

Question 15. (1/(x - 3)) + (2/(x - 2)) = 8/x, x ≠ 0, 2, 3

Solution:

 We have equation,

(1/(x - 3)) + (2/(x - 2)) = 8/x

(x - 2 + 2x - 6)/((x - 3) (x - 2)) = 8/x

(3x - 8)/((x - 3) (x - 2)) = 8/x 

8 ((x - 3) (x - 2)) = x(3x - 8)

5x² - 32x + 48 = 0

We can factorize this equation as:

5x² - 20x -12x + 48 = 0

5x (x - 4) -12(x - 4) = 0

(5x - 12) (x - 4) = 0

Therefore, the roots of equations are 12/5 or 4.

Question 16. a²x² - 3abx + 2b² = 0

Solution:

 We have equation,

a²x² - 3abx + 2b² = 0

We can factorize this equation as:

a²x² - 2abx - abx + 2b² = 0

ax (ax - 2b) - b (ax - 2b) = 0

(ax - b) (ax - 2b) = 0

Therefore, the roots of the equation are b/a or 2b/a. 

Question 17. 9x² - 6b²x - (a⁴ - b⁴) = 0

Solution:

 We have equation,

9x² - 6b²x - (a⁴ - b⁴) = 0

We can factorize this equation as:

(3x)² - 6b²x + (b²)² - a⁴ = 0

(3x - b²)² - (a²)² = 0

(3x - b² + a²) (3x - b² - a²) = 0

Therefore, the roots of the equation are (b² - a²)/3 or (b² + a²)/3. 

Question 18. 4x² + 4bx - (a² - b²) = 0

Solution:

We have equation,

4x² + 4bx - (a² - b²) = 0

Dividing by 4

x² + bx - ((a² - b²)/4) = 0

x² + bx - (((a - b)/2) ((a + b)/2)) = 0

We can write b as :

b = ((a + b)/2) - ((a - b)/2)

Therefore,

x² + ((a + b)/2) - ((a - b)/2) x - (((a - b)/2) ((a + b)/2)) = 0

x (x + ((a + b)/2)) - ((a - b)/2) (x + (a + b)/2) = 0

(x + ((a + b)/2)) (x - ((a - b)/2)) = 0

Therefore, the roots of the equation are (a - b)/2 or (-a - b)/2. 

Question 19. ax² + (4a² - 3b) x - 12ab = 0.

Solution:

We have equation,

 ax² + 4a²x - 3bx - 12ab = 0

ax (x + 4a) - 3b (x - 4a) = 0

(ax - 3b) (x + 4a) = 0

Therefore, the roots of the equation are 3b/a or -4a.

Question 20. 2x² + ax - a² = 0.

Solution:

We have equation,

2x² + ax - a² = 0

2x² + 2ax - ax - a² = 0

2x (x + a) - a (x + a) = 0

(2x - a) (x + a) = 0

Therefore, the roots of the equation are a/2 or -a.