# 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. 3x^{2 }– 14x – 5 = 0

**Solution:**

We have equation,

3x

^{2}– 14x – 5 = 0We can factorize this equation as

3x

^{2}– 15x + x – 5 = 03x (x – 5) -1 (x – 5) = 0

(3x – 1) (x – 5) = 0

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

### Question 4. 9x^{2 }– 3x – 2 = 0

**Solution:**

We have equation,

9x

^{2}– 3x – 2 = 0We can factorize this equation as

9x

^{2 }– 6x + 3x – 2 = 03x(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

x

^{2}+ 4x – 12 = 0We can factorize this equation as

x

^{2}+ 6x – 2x – 12 = 0x (x + 6) – 2 (x + 6) = 0

(x – 2) (x + 6) = 0

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

### Question 6. 6x^{2} + 11x + 3 = 0

**Solution:**

We have equation,

6x

^{2 }+ 11x + 3 = 0We can factorize this equation as

6x

^{2 }+ 9x + 2x + 3 = 03x (2x + 3) +1 (2x + 3) = 0

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

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

### Question 7. 5x^{2} – 3x – 2 = 0

**Solution:**

We have equation,

5x

^{2}– 3x – 2 = 0We can factorize this equation as

5x

^{2}– 5x + 2x – 2 = 05x (x – 1) + 2(x – 1) = 0

(x – 1) (5x + 2) = 0

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

### Question 8. 48x^{2} – 13x – 1 = 0.

**Solution:**

We have equation,

48x

^{2}– 13x – 1 = 0We can factorize this equation as

48x

^{2}– 16x + 3x – 1 = 016x (3x – 1) + 1 (3x – 1) = 0

(3x – 1) (16x + 1) = 0

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

### Question 9. 3x^{2} = -11x – 10.

**Solution:**

We have equation,

3x

^{2}+ 11x +10 = 0We can factorize this equation as

3x

^{2}+ 6x + 5x + 10 = 03x (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,

25x

^{2}+ 25x + 4 = 0We can factorize this equation as

25x

^{2}+ 20x + 5x + 4 = 05x (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

16x^{2}-27x -10 = 0We can factorize this equation as:

16x

^{2}– 32x + 5x – 10 = 016x (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

3x^{2}– 6x + 2 = 0Here, a = 3, b = -6 and c = 2

Since,

Discriminant (D) = b

^{2}– 4ac and x = (-b ± √D)/2aTherefore,

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

x^{2}– 3x -1 = 0Here, a = 1, b = -3 and c = -1

Since,

Discriminant (D) = b

^{2}– 4ac and x = (-b ± √D)/2aTherefore,

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

x

^{2}– 3x + 2 = 0We can factorize this equation as:

x

^{2}– 2x – x + 2 = 0x (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)

5x

^{2}– 32x + 48 = 0We can factorize this equation as:

5x

^{2}– 20x -12x + 48 = 05x (x – 4) -12(x – 4) = 0

(5x – 12) (x – 4) = 0

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

### Question 16. a^{2}x^{2} – 3abx + 2b^{2} = 0

**Solution:**

We have equation,

a

^{2}x^{2}– 3abx + 2b^{2}= 0We can factorize this equation as:

a

^{2}x^{2}– 2abx – abx + 2b^{2}= 0ax (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. 9x^{2} – 6b^{2}x – (a^{4} – b^{4}) = 0

**Solution:**

We have equation,

9x

^{2}– 6b^{2}x – (a^{4}– b^{4}) = 0We can factorize this equation as:

(3x)

^{2}– 6b^{2}x + (b^{2})^{2}– a^{4}= 0(3x – b

^{2})^{2}– (a^{2})^{2}= 0(3x – b

^{2 }+ a^{2}) (3x – b^{2}– a^{2}) = 0Therefore, the roots of the equation are

(b^{2}– a^{2})/3 or (b^{2}+ a^{2})/3.

### Question 18. 4x^{2} + 4bx – (a^{2 }– b^{2}) = 0

**Solution:**

We have equation,

4x

^{2}+ 4bx – (a^{2}– b^{2}) = 0Dividing by 4

x

^{2}+ bx – ((a^{2}– b^{2})/4) = 0x

^{2}+ bx – (((a – b)/2) ((a + b)/2)) = 0We can write b as :

b = ((a + b)/2) – ((a – b)/2)

Therefore,

x

^{2}+ ((a + b)/2) – ((a – b)/2) x – (((a – b)/2) ((a + b)/2)) = 0x (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. ax^{2} + (4a^{2} – 3b) x – 12ab = 0.

**Solution:**

We have equation,

ax

^{2}+ 4a^{2}x – 3bx – 12ab = 0ax (x + 4a) – 3b (x – 4a) = 0

(ax – 3b) (x + 4a) = 0

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

### Question 20. 2x^{2} + ax – a^{2} = 0.

**Solution:**

We have equation,

2x

^{2}+ ax – a^{2 }= 02x

^{2}+ 2ax – ax – a^{2}= 02x (x + a) – a (x + a) = 0

(2x – a) (x + a) = 0

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