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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.3 | Set 1
  • Last Updated : 18 Mar, 2021

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.

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