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

RD Sharma Class 10 Solutions Chapter 8 Ex 8.3 PDF Free Download

A quadratic equation can be solved in many ways. The solution of a quadratic equation by factorization method is the key concept discussed in this exercise. Students facing difficulties in any chapter can access the RD Sharma Solutions Class 10 for free to build up strong knowledge in the subject. The RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations Exercise 8.3 PDF is also provided below to guide students when solving it.

RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations Exercise 8.3 Download PDF

RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 10
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 11
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 12
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 13
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 14
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 15
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 16
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 17
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 18
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 19
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 20
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 21
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 22
RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations 23

Access RD Sharma Solutions for Class 10 Chapter 8 Quadratic Equations Exercise 8.3

Solve the following quadratic equation by factorization:

1. (x – 4)(x + 2) = 0

Solution:

Given,

(x – 4) (x + 2) = 0

So, either x – 4 = 0 ⇒ x = 4

Or, x + 2 = 0, ⇒ x = – 2

Thus, the roots of the given quadratic equation are 4 and -2 respectively. 

2. (2x + 3) (3x – 7) = 0

Solution:

Given,

(2x + 3) (3x – 7) = 0.

So, either 2x + 3 = 0, ⇒ x = – 3/2  

Or, 3x -7 = 0, ⇒ x = 7/3 

Thus, the roots of the given quadratic equation are x = -3/2 and x = 7/3 respectively. 

 

3. 3x– 14x – 5 = 0

Solution:

Given.

3x– 14x – 5 = 0

⇒ 3x– 14x – 5 = 0

⇒ 3x– 15x + x – 5 = 0

⇒ 3x(x – 5) + 1(x – 5) = 0

⇒ (3x + 1)(x – 5) = 0

Now, either 3x + 1 = 0 ⇒ x = -1/3  

Or, x – 5 = 0 ⇒ x = 5

Thus, the roots of the given quadratic equation are 5 and x = – 1/3 respectively. 

4. Find the roots of the equation 9x– 3x – 2 = 0.

Solution:

Given,

9x– 3x – 2 = 0.

⇒ 9x– 3x – 2 = 0.

⇒ 9x2 – 6x + 3x – 2 = 0

⇒ 3x (3x – 2) + 1(3x – 2) = 0

⇒ (3x – 2)(3x + 1) = 0

Now, either 3x – 2 = 0 ⇒ x = 2/3

Or, 3x + 1= 0 ⇒ x = -1/3 

Thus, the roots of the given quadratic equation are x = 2/3 and x = -1/3 respectively.   

 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 1

5.

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2

Dividing by 6 both the sides and cross-multiplying we get

x2+ 4x – 12 = 0

⇒ x+ 6x – 2x – 12 = 0

⇒ x(x + 6) – 2(x – 6) = 0

⇒ (x + 6)(x – 2) = 0

Now, either x + 6 = 0 ⇒x = -6

Or, x – 2 = 0 ⇒ x = 2

Thus, the roots of the given quadratic equation are 2 and – 6 respectively.

6. 6x+ 11x + 3 = 0

Solution:

Given equation is 6x+ 11x + 3 = 0.

⇒ 6x+ 9x + 2x + 3 = 0

⇒ 3x (2x + 3) + 1(2x + 3) = 0

⇒ (2x +3) (3x + 1) = 0

Now, either 2x + 3 = 0 ⇒ x = -3/2

  Or, 3x + 1= 0 ⇒ x = -1/3 

Thus, the roots of the given quadratic equation are x = -3/2 and x = -1/3 respectively. 

7. 5x– 3x – 2 = 0

Solution:

Given equation is 5x– 3x – 2 = 0.

⇒ 5x– 3x – 2 = 0.

⇒ 5x2 – 5x + 2x – 2 = 0

⇒ 5x(x – 1) + 2(x – 1) = 0

⇒ (5x + 2)(x – 1) = 0

Now, either 5x + 2 = 0 ⇒x = -2/5

Or, x -1= 0 ⇒x = 1

Thus, the roots of the given quadratic equation are 1 and x = -2/5 respectively. 

 

8. 48x– 13x – 1 = 0

Solution:

Given equation is 48x– 13x – 1 = 0.

⇒ 48x– 13x – 1 = 0.

⇒ 48x– 16x + 3x – 1 = 0.

⇒ 16x(3x – 1) + 1(3x – 1) = 0

⇒ (16x + 1)(3x – 1) = 0

Either 16x + 1 = 0 ⇒ x = -1/16  

Or, 3x – 1=0 ⇒ x = 1/3 

Thus, the roots of the given quadratic equation are x = -1/16 and x = 1/3 respectively. 

 

9. 3x= -11x – 10 

Solution:

Given equation is 3x= -11x – 10

⇒ 3x+ 11x + 10 = 0

⇒ 3x+ 6x + 5x + 10 = 0

⇒ 3x(x + 2) + 5(x + 2) = 0

⇒ (3x + 2)(x + 2) = 0

Now, either 3x + 2 = 0 ⇒ x = -2/3

Or, x + 2 = 0 ⇒ x = -2

Thus, the roots of the given quadratic equation are x = -2/3 and -2 respectively. 

 

10. 25x(x + 1) = – 4 

Solution:

Given equation is 25x(x + 1) = -4

25x(x + 1) = -4

⇒ 25x+ 25x + 4 = 0

⇒ 25x+ 20x + 5x + 4 = 0

⇒ 5x (5x + 4) + 1(5x + 4) = 0

⇒ (5x + 4)(5x + 1) = 0

Now, either 5x + 4 = 0 therefore x = – 4/5

Or, 5x + 1 = 0 therefore x = -1 /5

Thus, the roots of the given quadratic equation are x = – 4/5 and x = -1/5 respectively. 

 

11. 16x – 10/x = 27

Solution:

Given,

16x – 10/x = 27

On multiplying x on both the sides we have,

⇒ 16x2 – 10 = 27x

⇒ 16x2 – 27x – 10 = 0

⇒ 16x2 – 32x + 5x – 10 = 0

⇒ 16x(x – 2) +5(x – 2) = 0

⇒ (16x + 5) (x – 2) = 0

Now, either 16x + 5 = 0 ⇒ x = -5/16

Or, x – 2 = 0 ⇒ x = 2

Thus, the roots of the given quadratic equation are x = – 5/16 and x = 2 respectively. 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 312.

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 4

On cross multiplying on both the sides we get,

2 = 3x(x – 2)

2 = 3x– 6x

3x2– 6x – 2 = 0

⇒ 3x2– 3x – 3x – 2 = 0

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 5

Now, either

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 6

Thus,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 7are the solutions of the given quadratic equations.

13. x – 1/x = 3, x ≠ 0

Solution:

Given,

x – 1/x = 3

On multiplying x on both the sides we have,

⇒ x2 – 1 = 3x

⇒ x2 – 3x – 1 = 0

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 8
R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 9

14.

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 10

Dividing by 11 both the sides and cross-multiplying we get,

⇒ x– 3x – 28 = – 30

⇒ x– 3x – 2 = 0

⇒ x– 2x – x – 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

Now, either x – 2 = 0 ⇒ x = 2

Or, x – 1 = 0 ⇒ x = 1

Thus, the roots of the given quadratic equation are 1 and 2 respectively. 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 11

15.

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 12

On cross multiplying we get,

⇒ x(3x – 8) = 8(x – 3)(x – 2)

⇒ 3x2 – 8x = 8(x2 – 5x + 6)

⇒ 8x2 – 40x + 48 – (3x2 – 8x) = 0

⇒ 5x2 – 32x + 48 = 0

⇒ 5x2 – 20x – 12x + 48 = 0

⇒ 5x(x – 4) – 12(x – 4) = 0

⇒ (x – 4)(5x – 12) = 0

Now, either x – 4 = 0 ⇒ x = 4

Or, 5x – 12 = 0 ⇒ x = 5/12

Thus, the roots of the given quadratic equation are 1 and 2 respectively. 

16. a2x– 3abx + 2b= 0

Solution:

Given equation is a2x– 3abx + 2b= 0

⇒ a2x– abx – 2abx + 2b= 0

⇒ ax(ax – b) – 2b(ax – b) = 0

⇒ (ax – b)(ax – 2b) = 0

Now, either ax – b = 0 ⇒ x = b/a

Or, ax – 2b = 0 ⇒ x = 2b/a   

Thus, the roots of the quadratic equation are x = 2b/a and x = b/a respectively. 

17. 9x2 – 6b2x – (a4 – b4) = 0

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 13

Thus, the roots of the quadratic equation are x = (b2 – a2)/3 and x = (a2 + b2)/3 respectively. 

18. 4x+ 4bx – (a– b2) = 0 

Solution:

Given,

4x+ 4bx – (a– b2) = 0 

For factorizing,

4(a– b2) = -4(a – b) (a + b) = [-2(a-b)] [2(a + b)]

⇒ 2(b – a)*2(b + a)

⇒ 4x2+ (2(b – a) + 2(b + a)) – (a – b)(a + b) = 0

So, now

4x2  + 2(b – a)x++ 2(b + a)x + (b – a)(a + b) = 0

⇒ 2x(2x + (b – a)) +(a + b)(2x + (b – a)) = 0

⇒ (2x + (b – a))(2x + b + a) = 0

Now, either (2x + (b – a)) = 0 ⇒x = (a – b)/2

Or, (2x + b + a) = 0 ⇒ x = -(a + b)/2   

Thus, the roots of the given quadratic equation are x = -(a + b)/2 and x = (a – b)/2 respectively.

19. ax+ (4a– 3b)x – 12ab = 0

Solution:

Given equation is ax+ (4a– 3b)x – 12ab = 0

⇒ ax+ 4a2x – 3bx – 12ab = 0

⇒ ax(x + 4a) – 3b(x + 4a) = 0

⇒ (x + 4a)(ax – 3b) = 0

Now, either x + 4a = 0 ⇒ x = -4a

Or, ax – 3b = 0 ⇒ x = 3b/a 

Thus, the roots of the given quadratic equation are x = 3b/a and -4a respectively.

20. 2x2 + ax – a2 = 0

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 14

Thus, the roots of the given quadratic equation are x = a/2 and -a respectively.

21. 16/x – 1 = 15/(x + 1), x ≠ 0, -1

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 15

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 16

Thus, the roots of the given quadratic equation are x = 4 and -4 respectively.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 17

22. , x ≠ -2, 3/2

Solution:

Given,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 18

On cross-multiplying we get,

(x + 3)(2x – 3) = (x + 2)(3x – 7)

⇒ 2x– 3x + 6x – 9 = 3x– x – 14

⇒ 2x+ 3x – 9 = 3x– x – 14

⇒ x– 3x – x – 14 + 9 = 0

⇒ x– 5x + x – 5 = 0

⇒ x(x – 5) + 1(x – 5) = 0

⇒ (x – 5)(x + l) – 0

Now, either x – 5 = 0 or x + 1 = 0

⇒ x = 5 and x = -1

Thus, the roots of the given quadratic equation are 5 and -1 respectively. 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 1923.

, x ≠ 3, 4

Solution:

The given equation is

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 20

On cross multiplying, we have

3(4x– 19x + 20) = 25(x– 7x + 12)

⇒ 12x– 57x + 60 = 25x2 – 175x + 300

⇒13x– 78x – 40x + 240 = 0

⇒13x– 118x + 240 = 0

⇒13x– 78x – 40x + 240 = 0

⇒13x(x – 6) – 40(x – 6) = 0

⇒ (x – 6)(13x – 40) = 0

Now, either x – 6 = 0 ⇒x = 6

Or, 13x – 40 = 0 ⇒x = 40/13 

Thus, the roots of the given quadratic equation are 6 and 40/13 respectively. 

 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2124. x ≠ 0, 2

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 22

On cross multiplying, we get

4(2x+ 2) = 17(x– 2x)

⇒ 8x+ 8 = 17x– 34x

⇒ 9x– 34x – 8 = 0

⇒ 9x– 36x + 2x – 8 = 0

⇒ 9x(x – 4) + 2(x – 4) = 0

⇒ 9x + 2)(x – 4) = 0

Now, either 9x + 2 = 0 ⇒x = -2/9

Or, x – 4 = 0 ⇒ x = 4

Thus, the roots of the given quadratic equation are x = -2/9 and 4 respectively. 

 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2325.

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 24

On cross-multiplying, we get

7(-12x) = 48(x2 – 9)

⇒ -84x = 48x2 – 432

⇒ 48x2 + 84x – 432 = 0

⇒ 4x2 + 7x – 36 = 0 dividing by 12]

⇒ 4x2 + 16x – 9x – 36 = 0

⇒ 4x(x + 4) – 9(x – 4) = 0

⇒ (4x – 9)(x + 4) = 0

Now, either 4x – 9 = 0 ⇒x = 9/4

Or, x + 4 = 0 ⇒ x = -4

Thus, the roots of the given quadratic equation are x = 9/4 and -4 respectively. 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2526.

, x ≠ 0

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 26
R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 27

On cross multiplying, we have

x(3x – 5) = 6(x– 3x + 2)

⇒ 3x– 5x = 6x– 18x + 12

⇒ 3x– 13x + 12 = 0

⇒ 3x– 9x – 4x + 12 = 0

⇒ 3x(x – 3) – 4(x – 3) = 0

⇒ (x – 3)(3x – 4) = 0

Now, either x – 3 = 0 ⇒ x = 3

Or, 3x – 4 = 0 ⇒ 4/3.

Thus, the roots of the given quadratic equation are 3 and 4/3 respectively.

 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 2827.

, x ≠ 1, -1

Solution:

The given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 29

On cross – multiplying we have,

⇒ 6(4x) = 5(x– 1) = 24x

⇒ 5x– 5 = 5x– 24x – 5 =0

⇒ 5x– 25x + x – 5 = 0

⇒ 5x(x – 5) + 1(x – 5) = 0

⇒ (5x + 1)(x – 5) = 0

Now, either x – 5 = 0 ⇒ x = 5

Or, 5x + 1 = 0 ⇒ x = −1/5 

Thus, the roots of the given quadratic equation are x = −1/5 and 5 respectively. 

 

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 3028.

, x ≠ 1, -1/2

Solution:

The given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 31

On cross – multiplying we have,

⇒ 2(5x+ 2x + 2) = 5(2x– x – 1)

⇒ 10x+ 4x + 4 = 10x– 5x – 5

⇒ 4x + 5x + 4 + 5 = 0

⇒ 9x + 9 = 0

⇒ 9x = -9

Thus, x = -1 is the only root of the given equation.

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 3229.

 

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 33

Thus, the roots of the given quadratic equation are x = 1 and x = -2 respectively.

30.

Solution:

Given equation is,

R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.3 - 34

On cross-multiplying we have,

3(2x2 – 22x + 58) = 10(x2 – 12x + 35)

⇒ 6x2 – 66x + 174 = 10x2 – 120x + 350

⇒ 4x2 – 54x + 176 = 0

⇒ 2x2 – 27x + 88 = 0

⇒ 2x2 – 16x – 11x + 88 = 0

⇒ 2x(x – 8) – 11(x + 8) = 0

⇒ (x – 8)(2x – 11) = 0

Now, either x – 8 = 0 ⇒ x = 8

Or, 2x – 11 = 0 ⇒ x = 11/2 

Thus, the roots of the given quadratic equation are x = 11/2 and 8 respectively. 

Leave a Comment

Your email address will not be published. Required fields are marked *