Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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, 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

(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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!