# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.3 | Set 2

### Question 21. (16/x) – 1 = 15/(x + 1), x â‰  0, -1.

Solution:

We have equation,

(16/x) – 1 = 15/(x + 1)

(16 – x)/x = 15/(x + 1)

15x = (x + 1) (16 – x)

15x = 16x – x2 + 16 – x

15x – 16x + x2 -16 + x = 0

x2 – 16 = 0

(x – 4) (x + 4) = 0

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

Question 22. (x + 3)/(x + 2) = (3x – 7)/(2x – 3), x â‰  -2, 3/2.

Solution:

We have equation,

(x + 3) (2x – 3) = (3x – 7) (x + 2)

2x2 – 3x + 6x – 9 = 3x2 + 6x -7x -14

2x2 + 3x -9 = 3x2 -x -14

x2 -4x -5 = 0

We can factorize this equation as:

x2 – 5x + x -5 = 0

x (x – 5) + 1 (x – 5) = 0

(x + 1) (x – 5) = 0

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

### Question 23. (2x/(x – 4)) + ((2x – 5)/(x – 3)) = 25/3, x â‰  3, 4

Solution:

We have equation,

(2x/(x – 4)) + ((2x – 5)/(x – 3)) = 25/3

((2x)(x – 3) + (2x – 5) (x – 4))/((x – 4) (x – 3)) = 25/3

25x2 – 175x + 300 = 12x2 – 57x + 60

13x2 – 118x + 240 = 0

We can factorize this equation as:

13x2 – 78x – 40x + 240 = 0

13x (x – 6) – 40 (x – 6) = 0

(13x – 40) (x – 6) = 0

Therefore, roots of the equation are 6 or 40/13.

### Question 24. ((x + 3)/(x – 2)) – ((1 – x)/(x)) = 17/4, x â‰  0, 2

Solution:

We have equation,

((x + 3)/(x – 2)) + ((1 – x)/(x)) = 17/4

8x2 + 8 = 17x2 – 34x

9x2 – 34x – 8 = 0

We can factorize this equation as:

9x2 – 36x + 2x – 8 = 0

9x(x – 4) + 2(x – 4) = 0

(9x + 2) (x – 4) = 0

Therefore, roots of the equation are 4 or -2/9.

### Question 25. ((x – 3)/(x + 3)) – ((x + 3)/(x – 3)) = 48/7, x â‰  3, -3

Solution:

We have equation,

((x – 3)/(x + 3)) – ((x + 3)/(x – 3)) = 48/7

-84x = 48x2 – 432

48x2 + 84x – 432 = 0

4x2 + 7x -36 = 0

We can factorize this equation as:

4x2 + 16x – 9x – 36 = 0

4x (x + 4) – 9 (x + 4) = 0

(4x – 9) (x + 4) = 0

Therefore, roots of the equation are -4 or 9/4.

### Question 26. (1/(x – 2)) + (2/(x – 1)) = 6/x, x â‰  0

Solution:

We have equation,

(1/(x – 2)) + (2/(x – 1)) = 6/x

3x2 – 5x = 6x2 – 18x + 12

3x2 – 13x + 12 = 0

We can factorize this equation as:

3x2 – 9x -4x + 12 = 0

3x (x – 3) – 4 (x – 3) = 0

(3x – 4) (x – 3) = 0

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

### Question 27. ((x + 1)/(x – 1)) – ((x – 1)/(x + 1)) = 5/6, x â‰  1, -1

Solution:

We have equation,

((x + 1)/(x – 1)) – ((x – 1)/(x + 1)) = 5/6

5x2 – 5 = 24x

5x2 – 24x – 5 = 0

We can factorize this equation as:

5x2 – 25x + x – 5 = 0

5x (x – 5) + 1 (x – 5) = 0

(5x + 1) (x – 5) = 0

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

### Question 28. ((x – 1)/(2x + 1)) + ((2x + 1)/(x – 1)) = 5/2, x â‰  1, -1/2

Solution:

We have equation,

((x – 1)/(2x + 1)) + ((2x + 1)/(x – 1)) = 5/2

2 (5x2 + 2x + 2) = 5 (2x2 – x – 1)

9x + 9 = 0

9 (x + 1) = 0

Therefore, roots of the equation are -1.

### Question 29. (4/x) – 3 = 5/(2x + 3), x â‰  0, -3/2

Solution:

We have equation,

(4/x) – 3) = 5/(2x + 3)

5x = (2x + 3) (4 – 3x)

5x = 8x – 6x2 + 12 – 9x

6x2 – 6x -12 = 0

x2 – x -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, roots of the equation are 1 or -2.

### Question 30. ((x – 4)/(x – 5)) + ((x – 6)/(x – 7)) = 10/3, x â‰  5, 7

Solution:

We have equation,

((x – 4)/(x – 5)) + ((x – 6)/(x – 7)) = 10/3

4x2 – 54x + 176 = 0

2x2 – 27x + 88 = 0

We can factorize this equation as:

2x2 – 16x -11x + 88 = 0

2x (x – 8) – 11 (x – 8) = 0

(2x – 11) (x – 8) = 0

Therefore, roots of the equation are 8 or 11/2.

### Question 31. ((x – 2)/(x – 3)) + ((x – 4)/(x – 5)) = 10/3, x â‰  3, 5.

Solution:

We have equation,

((x – 2)/(x – 3)) + ((x – 4)/(x – 5)) = 10/3

We can rewrite it as :

((x – 3 + 1)/(x – 3)) + ((x – 5 + 1)/(x – 5)) = 10/3

1 + 1 + (1/(x – 3)) + (1/( x – 5)) = 10 3

4 (x2 – 8x + 15) = 6x – 24

4x2 – 38x + 84 = 0

2x2 -19x + 42 = 0

We can factorize this equation as:

2x2 – 12x – 7x + 42 = 0

2x (x – 6) – 7 (x – 6) = 0

(2x – 7) (x – 6) = 0

Therefore, roots of the equation are 6 or 7/2.

### Question 32. (( 5 + x)/(5 – x)) – ((5 – x)/(5 + x)) = 15/4, x â‰  5, -5.

Solution:

We have equation,

(( 5 + x)/(5 – x)) – ((5 – x)/(5 + x)) = 15/4

80x = 375 – 15x2

15x2 +80x -375 = 0

3x2 +16x – 75 = 0

We can factorize this equation as:

3x2 + 25x – 9x – 75 = 0

x (3x + 25) – 3 (3x + 25) = 0

(3x + 25) (x – 3) = 0

Therefore, roots of the equation are 3 or -25/3.

### Question 33. (3/(x + 1)) – (1/2) = 2/(3x – 1), x â‰  -1, 1/3

Solution:

We have equation,

(3/(x + 1)) – (1/2) = 2/(3x – 1)

2 (2x + 2) = (5 – x) (3x – 1)

4x + 4 = 15x – 5 – 3x2  + x

3x2 – 12x + 9 = 0

x2 – 4x + 3 = 0

We can factorize this equation as:

x2 – 3x – x + 3 = 0

x (x – 3) – 1 (x – 3) = 0

(x – 3) (x – 1) = 0

Therefore, roots of the equation are 1 or 3.

### Question 34. (3/(x + 1)) + (4/(x – 1)) = 29/(4x – 1), x â‰  -1, 1, 1/4

Solution:

We have equation,

(3/(x + 1)) + (4/(x – 1)) = 29/(4x – 1)

(7x + 1) (4x – 1) = 29 (x2 – 1)

28x2 – 7x + 4x – 1 = 29x2 – 29

x2 + 3x -28 = 0

We can factorize this equation as:

x2 + 7x – 4x – 28 = 0

x (x + 7) – 4 (x + 7) = 0

(x – 4) (x + 7) = 0

Therefore, roots of the equation are 4 or -7.

### Question 35. (2/(x + 1)) + (3/(2(x – 2))) = 23/5x, x â‰  0, -1, 2

Solution:

We have equation,

(2/(x + 1)) + (3/(2(x – 2))) = 23/5x

35x2 – 25x = 46 (x2 – x – 2)

11x2 – 21x – 92 = 0

We can factorize this equation as:

11x2 – 44x + 23x – 92 = 0

11x (x – 4) + 23 (x – 4) = 0

(11x + 23) ( x – 4) = 0

Therefore, roots of the equation are 4 or -23/11.

### Question 36.  x2 – (âˆš3 + 1) x + âˆš3 = 0

Solution:

We have equation,

x2 – (âˆš3 + 1) x + âˆš3 = 0

x2 – âˆš3x – x + âˆš3 = 0

We can factorize this equation as:

x (x – âˆš3) – 1 (x – âˆš3) = 0

(x -1) (x – âˆš3) = 0

Therefore, roots of the equation are 1 or âˆš3.

### Question 37. 3âˆš5 x2 + 25x – 10âˆš5 = 0

Solution:

We have equation,

3âˆš5 x2 + 25x – 10âˆš5 = 0

âˆš5 (3x2 + (25/âˆš5) x – (10âˆš5/âˆš5)) = 0

âˆš5 (3x2 + 5âˆš5x – 10) = 0

We can factorize this equation as:

3x2 – âˆš5x + 6âˆš5x – 10 = 0

x (3x – âˆš5) + 2âˆš5 (3x – âˆš5) = 0

(x + 2âˆš5) (3x – âˆš5) = 0

Therefore, roots of the equation are -2âˆš5 or âˆš5/3.

### Question 38. âˆš3x2 – 2âˆš2 x – 2âˆš3 = 0

Solution:

We have equation,

âˆš3x2 – 2âˆš2 x – 2âˆš3 = 0

Here a = âˆš3, b = -2âˆš2 and c = -2âˆš3

Since, Discriminant D = b2 – 4ac and x = (-b Â± âˆšD)/2a

Therefore,

D = 8 + 24 = 32, and

x = (-(-2âˆš2) Â± âˆš32)/2âˆš3

x = ( 2âˆš2 Â± 4âˆš2)/2âˆš3

Therefore, roots of the equations are (2âˆš2 + 4âˆš2)/2âˆš3 or (2âˆš2 – 4âˆš2)/2âˆš3.

### Question 39. 4âˆš3x2 + 5x – 2âˆš3 = 0

Solution:

We have equation,

4âˆš3x2 + 5x – 2âˆš3 = 0

Here a = 4âˆš3, b = 5 and c = -2âˆš3

Since, Discriminant D = b2 – 4ac and x = (-b Â± âˆšD)/2a

Therefore,

D = 25 + 96 = 121, and

x = (-(5) Â± âˆš121)/8âˆš3

x = (-5 Â± 11)/8âˆš3

Therefore, roots of the equations are -2 /âˆš3 or âˆš3/ 4.

### Question 40. âˆš2x2 – 3x – 2âˆš2 = 0

Solution:

We have equation,

âˆš2x2 – 3x – 2âˆš2 = 0

Here a = âˆš2, b = -3 and c = -2âˆš2

Since, Discriminant D = b2 – 4ac and x = (-b Â± âˆšD)/2a

Therefore,

D = 9 + 16 = 25, and

x = (-(-3) Â± âˆš25)/2âˆš2

x = (3 Â± 5)/2âˆš2

Therefore, roots of the equations are 2âˆš2 or -1/ âˆš2.

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