# RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10A Class 10 Maths

 Chapter Name RS Aggarwal Chapter 10 Quadratic Equation Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 10BExercise 10CExercise 10DExercise 10EExercise 10F Related Study NCERT Solutions for Class 10 Maths

### Exercise 10A Solutions

1. Which of the following are quadratic equation in x ?

(i) x2 – x + 2 = 0

(ii) 2x2 + 5/2x – 3 = 0

(iii) √2x2 + 7x + 5√2

(iv) 1/3x2 + 1/5x – 2 = 0

(v) x2 – 3x - √x + 4 = 0

(vi) x – 6/x = 3

(vii) x2 – 2/x = x2

(viii) x2 – 1/x2 = 5

(ix) (x + 2)3 = x3 – 8

(x) (2x + 3(3x + 2) = 6(x – 1)(x – 2)

(xi) (x + 1/x)2 = 2(x + 1/x) + 3

Solution

(i) (x2 – x + 3) is a quadratic polynomial

∴ x2 – x + 3 = 0 is a quadratic equation.

(ii) Clearly, (2x2 + 5/2.x - √3) is a quadratic polynomial.

∴ 2x2 + 5/2.x - √3 = 0 is a quadratic equation.

(iii) Clearly, (√2x2 + 7x + 5√2) is a quadratic polynomial.

√2x2 + 7x + 5√2 = 0 is a quadratic equation.

(iv) Clearly, (1/3x2 + 1/5.x – 2) is a quadratic polynomial.

∴ 1/3x2 + 1/5x – 2 = 0 is a quadratic equation.

(v) (x2 – 3x - √x + 4) contains a term with √x, i.e., x1/2, where 1/2 is not a integer.

Therefore, it is not a quadratic polynomial.

∴ x2 – 3x - √x + 4 = 0 is not a quadratic equation.

(vi) x – 6/x = 3

⇒ x2 – 6 = 3x

⇒ x2 – 3x – 6 = 0

(x3 – 3x – 6) is not quadratic polynomial; therefore, the given equation is quadratic.

(vii) x2 – 2/x = x2

⇒ x2 + 2 = x3

⇒ x3 – x2 – 2 = 0

(x3 – x2 – 2) = 0 is not a quadratic polynomial.

∴ x3 – x2 – 2 = 0 is not a quadratic equation.

(viii) x2 – 1/x2 = 5

⇒ x4 – 1 = 5x2

⇒ x4 – 5x2 – 1 = 0

(x4 – 5x2 - 1) is a polynomial with degree 4.

∴ x4 – 5x2 – 1 = 0 is not a quadratic equation.

(ix) (x + 2)3 = x3 – 8

⇒ x3 + 6x2 + 12x + 8 = x3 – 8

⇒ 6x2 + 12x + 16 = 0

This is of form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(x) (2x + 3)(3x + 2) = 6(x – 1)(x – 2)

⇒ 6x2 + 4x + 9x + 6 = 6(x2 – 3x + 2)

⇒ 6x2 + 13x + 6 = 6x2 – 18x + 12

⇒ 31x – 6 = 0

This is of the form ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

(xi) (x + 1/x)2 = 2(x + 1/x) + 3

⇒ (x2 + 1)/x)2 = 2(x2 + 1)/x + 3

⇒ (x2 + 1)2 = 2x(x2 + 1) + 3x2

⇒ x4 + 2x2 + 1 = 2x3 + 2x + 3x2

⇒ x4 – 2x3 – x2 – 2x + 1 = 0

This is not of the form ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

2. Which of the following are the roots of 3x2 + 2x – 1 = 0 = ?

(i) – 1

(ii) 1/3

(iii) – 1/2

Solution

The given equation is (3x2 + 2x – 1 = 0).

(i) x = (-1)

LHS = x2 + 2x – 1

= 3 × (-1)2 + 2 × (-1) – 1

= 3 – 2 – 1

= 0

= R.H.S.

Thus, (-1) is a root of (3x2 + 2x – 1 = 0).

(ii) On subtracting x = 1/3 in the given equation, we get:

L.H.S. = 3x2 + 2x – 1

= 3 × (1/3)2 + 2 × 1/3 – 1

= 3 × 1/9 + 2/3 – 1

= (1 + 2 – 3)/3

= 0/3

= 0

= R.H.S.

Thus, (1/3) is a root of (3x2 + 2x – 1 = 0)

(iii) On subtracting x = (-1/2) in the given equation, we get

L.H.S = 3x2 + 2x – 1

= 3 × (-1/2)2 + 2 × (-1/2) – 1

= 3 × 1/4 – 1 – 1

= 3/4 – 2

= (3 – 8)/4

= -5/4 ≠ 0

Thus, L.H.S. = R.H.S.

Hence, (-1/2) is a solution of (3x2 + 2x – 1 = 0).

3. Find the value of k for which x = 1 is a root of the equation x2 + kx + 3 = 0.

Solution

It is given that (x = 1) is a root of (x2 + kx + 3 = 0).

Therefore, (x = 1) must satisfy the equation.

⇒ (1)2 + k × 1 + 3 = 0

⇒ k + 4 = 0

⇒ k = - 4

Hence, the required value of k is -4.

4. Find the value of a and b for which x = 3/4 and x = - 2 are the roots of the equation ax2 + bx – 6 = 0

Solution

It is given that 3/4 is a root of ax3 + bx – 6 = 0; therefore, we have:

a × (3/4)2 + b × 3/4 - 6 = 0

⇒ 9a/16 + 3b/4 = 6

⇒ (9a + 12b)/16 = 6

⇒ 9a + 12b – 96 = 0

⇒ 3a + 4b = 32 ....(i)

Again, (-2) is a root of ax2 + bx – 6 = 0; therefore, we have:

a × (-2)2 + b × (-2) – 6 = 0

⇒ 4a – 2b = 6

⇒ 2a – b = 3 ....(ii)

On multiplying (ii) by 4 and adding the result with (i), we get:

⇒ 3a + 4b + 8a – 4b = 32 + 12

⇒ 11a = 44

⇒ a = 4

Putting the value of a in (ii), we get:

2 × 4 – b = 3

⇒ 8 – b = 3

⇒ b = 5

Hence, the required values of a and b are 4 and 5, respectively.

5. (2x – 3)(3x + 1) = 0

Solution

(2x – 3)(3x + 1) = 0

⇒ 2x – 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = - 1

⇒ x = 3/2 or x = - 1/3

Hence, the roots of the given equation are 3/2 and -1/3.

6. 4x2 + 5x = 0

Solution

4x2 + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = - 5/4

Hence, roots of the given equations are 0 and -5/4.

7. 3x2 – 243 = 0.

Solution

Given:

3x2 – 243 = 0

⇒ 3(x2 – 81) = 0

⇒ (x)2 – (9)2 = 0

⇒ (x + 9)(x – 9) = 0

⇒ x + 9 = 0 or x – 9 = 0

⇒ x = - 9 or x = 9

Hence, -9 and 9 are the roots of the equation 3x2 – 243 = 0.

8. 2x2 + x – 6 = 0

Solution

We write, x = 4x – 3 as 2x2 × (-6)

= -12x2

= 4x × (-3x)

∴ 2x2 + x – 6 = 0

⇒ 2x2 + 4x – 3x – 6 = 0

⇒ 2x(x + 2) – 3(x + 2) = 0

⇒ (x + 2)(2x – 3) = 0

⇒ x + 2 = 0 or 2x – 3 = 0

⇒ x = - 2 or x = 3/2

Hence, the roots of the given equations are -2 and 3/2.

9. x2 + 6x + 5 = 0

Solution

We write, 6x = x + 5x as x2 × 5 = 5x2 = x × 5x

∴ x2 + 6x + 5 = 0

⇒ x2 + x – 5x + 5 = 0

⇒ x(x + 1) + 5(x + 1) = 0

⇒ (x + 1)(x + 5) = 0

⇒ x + 1 = 0 or x + 5 = 0

⇒ x = - 1 or x = - 5

Hence, the roots of the given equations are -1 and -5.

10. 9x2 – 3x – 2 = 0

Solution

We write, -3x = 3x – 6x as 9x2 × (-2)

= -18x2

= 3x × (-6x)

∴ 9x2 – 3x – 2 = 0

⇒ 9x2 + 3x – 6x – 2 = 0

⇒ 3x(3x + 1) – 2(3x + 1) = 0

⇒ (3x + 1)(3x – 2) = 0

⇒ 3x + 1 = 0 or 3x – 2 = 0

⇒ x = -1/3 or x = 2/3

Hence, the roots of the given equations are -1/3 and 2/3.

11. x2+ 12x + 35 = 0

Solution

x2 + 12 + 35 = 0

⇒ x2 + 7x + 5x + 35 = 0

⇒ x(x + 7) + 5(x + 7) = 0

⇒ (x + 5)(x + 7) = 0

⇒ x + 5 = 0 or x + 7 = 0

⇒ x = - 5 or x = - 7

Hence, -5 and -7 are the roots of the equation x2 + 12x + 35 = 0.

12. x= 18x – 77

Solution

x2 = 18x – 77

⇒ x2 – 18x + 77 = 0

⇒ x2 – (11x + 7x) + 77 = 0

⇒ x2 – 11x – 7x + 77 = 0

⇒ x(x – 11) – 7(x – 11) = 0

⇒ (x – 7)(x – 11) = 0

⇒ x – 7 = 0 or x – 11 = 0

⇒ x = 7 or x = 11

Hence, 7 and 11 are the roots of the equations x2 = 18x – 77.

13. 6x2+ 11x + 3 = 0.

Solution

Given:

6x2 + 11x + 3 = 0

⇒ 6x2 + 9x + 2x + 3 = 0

⇒ 3x(2x + 3) + 1(2x + 3) = 0

⇒ (3x + 1)(2x + 3) = 0

⇒ 3x + 1 = 0 or 2x + 3 = 0

⇒ x = -1/3 or x = -3/2

Hence, -1/3 and -3/2 are the roots of the equation 6x2 + 11x + 3 = 0

14. 6x2+ x – 12 = 0.

Solution

Given:

6x2 + x – 12 = 0

⇒ 6x2 + 9x – 8x – 12 = 0

⇒ 3x(2x + 3) – 4(2x + 3) = 0

⇒ (3x – 4)(2x + 3) = 0

⇒ 3x – 4 = 0 or 2x + 3 = 0

⇒ x = 4/3 or x = -3/2

Hence, 4/3 and -3/2 are the roots of the equation 6x2 + x – 12 = 0.

15. 3x2 – 2x – 1 = 0

Solution

We write, -2x = -3x + x as 3x2 × (-1)

= -3x2

= (-3x) × x

∴ 3x2 – 2x – 1 = 0

⇒ 3x(x – 1) + 1(x – 1) = 0

⇒ (x – 1)(3x + 1) = 0

⇒ x – 1 = 0 or 3x + 1 = 0

⇒ x = 1 or x = -1/3

Hence, the roots of the given equation are 1 and – 1/3.

16. 4x2 – 9x = 100

Solution

Given:

4x2 – 9x = 100

⇒ 4x2 – 9x – 100 = 0

⇒ 4x2 – (25x – 16x) – 100 = 0

⇒ 4x2 – 25x + 16x – 100 = 0

⇒ x(4x – 25) + 4(4x – 25) = 0

⇒ (4x – 25)(x + 4) = 0

⇒ 4x – 25 = 0 x + 4 = 0

⇒ x = 25/4 or x = -4

Hence, the roots of the equation are 25/4 and -4.

17. 15x2 – 28 = x

Solution

Given:

15x2 – 28 = x

⇒ 15x2 – x – 28 = 0

⇒ 15x2 – (21x – 20x) – 28 = 0

⇒ 15x2 – 21x + 20x – 28 = 0

⇒ 3x(5x – 7) + 4(5x – 7) = 0

⇒ (3x + 4)(5x – 7) = 0

⇒ 3x + 4 = 0 or 5x – 7 = 0

⇒ x = -4/3 or x = 7/5

Hence, the roots of the equation are -4/3 and 7/5.

18. 4 – 11x = 3x2

Solution

Given:

4 – 11x = 3x2

⇒ 3x2 + 11x – 4 = 0

⇒ 3x2 + 12x – x – 4 = 0

⇒ 3x(x + 4) - 1(x + 4) = 0

⇒ (x + 4)(3x – 1) = 0

⇒ x + 4 = 0 or 3x – 1 = 0

⇒ x = -4 or x = 1/3.

1948x2 - 13x – 1 = 0

Solution

48x2 – 13x – 1 = 0

⇒ 48x2 – (16x – 3x) – 1 = 0

⇒ 48x2 – 16x + 3x – 1 = 0

⇒ 16x(3x – 1) + 1(3x – 1) = 0

⇒ (16x + 1)(3x – 1) = 0

⇒ 16x + 1 = 0 or 3x – 1 = 0

⇒ x = -1/16 or x = 1/3

Hence, the roots of equation are -1/16 and 1/3.

20. x2 + 2√2x – 6 = 0

Solution

2√2x = 3√2x - √2x as x2 × (-6)

= - 6x2

= 3√2x × (-√2x)

∴ x2 + 2√2x – 6 = 0

⇒ x2 + 2√2x - √2x – 6 = 0

⇒ x(x + 3√2) - √2(x + 3√2) = 0

⇒ (x + 3√2)(x - √2) = 0

⇒ x + 3√2 = 0 or x - √2 = 0

⇒ x = -3√2 or x = √2

Hence, the roots of the given equation are -3√2 and √2

21. √3x2 + 10x + 7√3 = 0

Solution

We write: 10x = 3x + 7x as √3x2 × 7√3

= 21x3

= 3x × 7x

∴ √3x2 + 10x + 7√3 = 0

⇒ √3x2 + 3x + 7x + 7√3 = 0

⇒ √3x(x + √3) + 7(x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

⇒ x + √3 = 0 or √3x + 7 = 0

⇒ x = -√3 or x = -7/√3 = -(7√3)/3

Hence, the roots of the given equation are -√3 and (-7√3)/3.

22.√3x2 + 11x + 6√3 = 0

Solution

√3x2 + 11x + 6√3 = 0

⇒ √3x2 + 9x + 2x + 6√3 = 0

⇒ √3x(x + 3√3) + 2(x + 3√3) = 0

⇒ (x + 3√3)(√3x + 2) = 0

⇒ x + 3√3 = 0 or √3x + 2 = 0

⇒ x = -3√3 or x = -2/√3

= (-2 × √3)/(√3 × √3)

= -2√3/3

Hence, the roots of the equation are -3√3 and (-2√3)/3.

23. 3√3 + 4x - √7 = 0

Solution

Given:

3√7x2 + 4x - √7 = 0

⇒ 3√7x2 + 7x – 3x - √7 = 0

⇒ √7x(3x + √7) – 1(3x + √7) = 0

⇒ (3x + √7)(√7 – 1) = 0

⇒ 3x + √7 = 0 or √7x – 1 = 0

⇒ x = -√7/3 or x = 1√7 = (1 × √7)/(√7 × √7)

= √7/7

Hence, the roots of the equations are -√7/3 and √7/7.

24. √7x2 – 6x - 13√7 = 0

Solution

We write, -6x = 7x – 13 as √7x2 × (-13√7)

= -91x2

= 7x × (-13x)

∴ √7x2 + 7x – 13x - 13√7 = 0

⇒ √7x2 + 7x – 13x - 13√7 = 0

⇒ √7x(x + √7) – 13(x + √7) = 0

⇒ (x + √7)(√7x – 13) = 0

⇒ x + √7 = 0 or √7x – 13 = 0

⇒ x = -√7 or x = 13/√7

= (13√7)/7

Hence, the roots of the given equation are - √7 and 13√7/7

25. 4√6x2 – 13x - 2√6 = 0

Solution

4√6x2 – 13x - 2√6 = 0

⇒ 4√6x2 – 16x + 3x - 2√6 = 0

⇒ 4√2x (√3x - 2√2) + √3(√3x - 2√2) = 0

⇒ (4√2x + √3)(√3x - 2√2) = 0

⇒ 4√2x + √3 = 0 or √3x - 2√2 = 0

⇒ x = -√3/(4√2)

= (-√3 × √2)/(4√2 × √2)

= -√8/8

or x = 2√2/√3

= (2√2 × √3)/(√3 × √3)

= (2√6)/3

Hence, the roots of the equations are -√6/8 and 2√6/3.

26. 3x2 - 2√6x + 2 = 0

Solution

We write, -2√6x

= -√6x and 3x2 × 2

= 6x2

= (-√6x) × (-√6x)

∴ 3x2 - 2√6x + 2 = 0

⇒ 3x2 - √6x - √6x + 2 = 0

⇒ √3x(√3x - √2) - √2(√3x - √2) = 0

⇒ (√3x - √2)(√3x - √2) = 0

⇒ (√3x - √2)2 = 0

⇒ √3x - √2 = 0

⇒ x = √2/√3 = √6/3

Hence, √6/3 vis the repeated root of the given equation.

27. √3x2 - 2√2x - 2√3 = 0

Solution

We write, -2√2x

= -3√2x + √2x as √3x2 × (-2√3)

= -6x2

= (-3√2x) × (√2x)

∴ √3x2 - 2√2x - 2√3 = 0

⇒ √3x2 - 3√2x + √2x - 2√3 = 0

⇒ √3x(x - √6) + √2(x - √6) = 0

⇒ (x - √6)( √3x + √2) = 0

⇒ x - √6 = 0 or √3x + √2 = 0

⇒ x - √6 = 0 or x = - √2/√3

= -√6/3

Hence, the roots of the given equation are √6 and -√6/3.

28. x2 - 3√5x + 10 = 0

Solution

We write, -3√5x

= -2√5x - √5x as x2 × 10 = 10x2

= (-2√5x) × (-√5x)

∴ x2 - 3√5x + 10 = 0

⇒ x2 - 2√5x - √5x + 10 = 0

⇒ x(x - 2√5) - √5(x - 2√5) = 0

⇒ (x - 2√5)(x - √5) = 0

⇒ x(x - 2√5) - √5(x - 2√5) = 0

Hence, the roots of the given equations are √5 and 2√5.

29. x2 – (√3 + 1)x + √3 = 0

Solution

x2 – (√3x + 1)x + √3 = 0

⇒ x2 - √3x – x + √3 = 0

⇒ x(x - √3) – 1(x - √3) = 0

⇒ (x - √3)(x – 1) = 0

⇒ x - √3 = 0 or x – 1 = 0

⇒ x = √3 or x = 1

Hence, 1 and √3 are the roots of the given equations.

30. x2 + 3√3x – 30 = 0

Solution

We write, 3√3x = 5√3x - 2√3x as x2 × (-30)

= -30x2

= 5√3x × (-2√3)x

∴ x2 + 3√3x – 30 = 0

⇒ x2 + 5√3x - 2√3x – 30 = 0

⇒ x(x + 5√3) - 2√3(x + 5√3) = 0

⇒ (x + 5√3)(x - 2√3) = 0

⇒ x + 5√3 = 0 or x - 2√3 = 0

⇒ x = -5/√3 or x = 2√3

Hence, the roots of the given equation are -5√3 and 2√3.

31. √2x2 + 7x + 5√2 = 0

Solution

We write, 7x = 5x + 2x as √2x2 × 5√2

= 10x2

= 5x × 2x

∴ √2x2 + 7x + 5√2 = 0

⇒ √2x2 + 5x + 2x + 5√2 = 0

⇒ x(√2x + 5) + √2(√2x + 5) = 0

⇒ (√2x + 5)(x + √2) = 0

⇒ x + √2 = 0 or √2x + 5 = 0

⇒ x = -√2 or x = -(5/√2) = -(5√2)/2

Hence, the roots of the given equation are -√2 and -(5√2)/2.

32. 5x2 + 13x + 8 = 0

Solution

We write, 13x = 5x + 8x as 5x2 × 8

= 40x2

= 5x × 8x

∴ 5x2 + 13x + 8 = 0

⇒ 5x2 + 5x + 8x + 8 = 0

⇒ 5x(x + 1) + 8(x +1) = 0

⇒ (x + 1)(5x + 8) = 0

⇒ x + 1 = 0 or 5x + 8 = 0

x = -1 or x = - (8/5)

Hence, -1 and -8/5 are the roots of the given equation.

33. x2 – (1 + √2)x + √2 = 0

Solution

Given:

x2 – (1 + √2)x + √2 = 0

⇒ x2 – x - √2x + √2 = 0

⇒ x(x – 1) - √2(x – 1) = 0

⇒ (x - √2)(x – 1) = 0

⇒ x - √2 = 0 or x – 1 = 0

⇒ x = √2 or x = 1

Hence, the roots of the equation are √2 and 1.

34. 9x2 + 6x + 1 = 0

Solution

9x2 + 6x + 1 = 0

⇒ 9x2 + 3x + 3x + 1 = 0

⇒ 3x(3x + 1) + 1(3x + 1) = 0

⇒ (3x + 1)(3x + 1) = 0

⇒ 3x + 1 = 0 or 3x + 1 = 0

⇒ x = -1/3 or x = -1/3

Hence, -1/3 is the root of the equation 9x2 + 6x + 1 = 0.

35. 100x2 – 20x + 1 = 0

Solution

We write, -20x = -10x – 10x as 100x2 × 1

= 100x2

= (-10x) × (-10x)

∴ 100x2 – 20x + 1 = 0

⇒ 100x2 – 10x – 10x + 1 = 0

⇒ 10x(10x – 1) - 1(10x – 1) = 0

⇒ (10x – 1)(10x – 1) = 0

⇒ (10x – 1)2 = 0

⇒ 10x – 1 = 0

⇒ x = 1/10

Hence, 1/10 is the repeated root of the given equation.

36. 2x2 – x + 1/8 = 0

Solution

We write, -x = -x/2 – x/2 as 2x2 × 1/8

= x2/4

= - (x/2) × - (x/2)

2x2 – x + 1/8 = 0

2x2 – x/2 – x/2 + 1/8 = 0

2x(x – 1/4) – 1/2(x – 1/4) = 0

(x – 1/4)(2x – 1/2) = 0

x – 1/4 = 0 or 2x – 1/2 = 0

x = 1/4 or x = 1/4

Hence, 1/4 is the repeated root of the given equation.

37. 10x – 1/x = 3

Solution

Given:

10x - 1/x = 3

10x2 – 1 = 3x  [Multiplying both sides by x]

10x2 – 3x – 1 = 0

10x2 – (5x – 2x) – 1 = 0

10x2 – 5x + 2x – 1 = 0

5x(2x – 1) + 1(2x – 1) = 0

(2x – 1)(5x + 1) = 0

2x – 1 = 0 or 5x + 1 = 0

x = 1/2 or x = -1/5

Hence, the roots of the equation are 1/2 and -1/5.

38. 2/x2 - 5/x + 2 = 0

Solution

Given:

2/x2 – 5/x + 2 = 0

2 – 5x + 2x2 = 0  [Multiplying both sides by x2]

2x2 – 5x + 2 = 0

2x2 – (4x + x) + 2 = 0

2x2 – 4x – x + 2 = 0

2x(x – 2) – 1(x – 2) = 0

(2x – 1)(x – 2) = 0

2x – 1 = 0 or x – 2 = 0

x = 1/2 or x = 2

Hence, the roots of the equation are 1/2 and 2.

39. 2x2 + ax – a2 = 0

Solution

We write, ax = 2ax – ax as 2x2 × (- a2)

= -2a2x2

= 2ax × (- ax)

∴ 2x2 + ax – a2 = 0

2x(x + a) – a(x + a) =- 0

(x + a)(2x – a) = 0

x + a = 0 or 2x – a = 0

x = -a or x = a/2

Hence, -a and a/2 are the roots of the given equation.

40. x2+ 4bx – (a2 – b2) = 0

Solution

We write, 4bx = 2(a + b)x – 2(a – b)x as

4x× [-(a2 – b2)]

= -4(a2 – b2)x2

= 2(a + b)x × [-2(a – b)x]

∴ 4x2 + 4bx – (a2 - b2) = 0

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

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

⇒ [2x + (a + b)][2x – (a – b)] = 0

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

⇒ x = -(a + b)/2 or x = (a – b)/2

Hence, -(a + b)/2 or x = (a – b)/2 are the roots of the given equation.

41. 4x2 - 4a2x + (a4 – b4) = 0

Solution

We write, -4a2x = -2(a2 + b2)x – 2(a2 – b2)x as

4x2 × (a4 – b4) = 4(a4 – b4)x2

= [-2(a2 + b2)]x × [-2(a2 – b2)]x

∴ 4x2 – 4a2x + (a4 – b4) = 0

⇒ 4x2 – 2(a2 + b2)x – 2(a2 – b2)x2 = [-2(a2 + b2)]x × [-2(a2 – b2)]x

⇒ 2x[2x – (a2 + b2)] – (a2 – b2)[2x – (a2 + b2)] = 0

⇒ [2x – (a2 + b2)] [2x – (a2 – b2)] = 0

⇒ 2x – (a2 + b2) = 0 or 2x – (a2 – b2) = 0

⇒ x = (a2 + b2)/2 or (a2 – b2)/2

Hence, (a2 + b2)/2 and (a2 – b2)/2 are the roots of the given equation.

42. x2 + 5x – (a2 + a – 6) = 0

Solution

We write, 5x = (a + 3)x – (a – 2)x as

x × [-(a2 + a – 6)]

= -(a2 + a – 6)x2

= (a + 3)x × [-(a – 2)x]

∴ x2 + 5x – (a2 + a – 6) = 0

⇒ x2 + (a + 3)x – (a – 2)x –(a + 3)(a – 2) = 0

⇒ x[x + (a + 3)] – (a – 2)[x + (a + 3)] = 0

⇒ [x + (a + 3)][x – (a – 2)] = 0

⇒ x + (a + 3) = 0, x - (a – 2) = 0

⇒ x = -(a + 3) or x = a – 2

Hence, -(a + 3) and (a – 2) are the roots of the given equation.

43. x2 – 2ax - (4b2 – a2) = 0

Solution

We have, -2ax = (2b – a)x – (2b + a)x as

x2 × [-(4b2 – a2)]

= -(4b2 – a2)x2

= (2b – a)x × [-(2b + a)x]

∴ x2 – 2ax – (4b2 – a2) = 0

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

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

⇒ [x + (2b – a)][x – (2b + a)] = 0

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

x = -(2b – a) or x = 2b + a

⇒ x = a – 2b or x = a + 2b

Hence, a – 2b and a + 2b are the roots of the given equation.

44. x2 – (2b – 1)x + (b2 – b – 20) = 0

Solution

We write, -(2b – 1)x = -(b – 5)x – (b + 4)x as

x2 × (b2 – b – 20) = (b2 – b – 20)x2

= [-(b – 5)x] × [-(b + 4)x]

∴ x2 – (2b – 1)x + (b2 – b – 20) = 0

⇒ x2 – (b – 5)x – (b + 4)x + (b – 5)(b + 4) = 0

⇒ x[x – (b – 5)] – (b + 4)[x – (b – 5)] = 0

⇒ [x – b – 5]9x + (b + 4)] = 0

⇒ x – (b – 5) = 0 or x – (b + 4) = 0

⇒ x = b – 5 or x = b + 4

Hence, b – 5 and b + 4 are the roots of the given equation.

45. x2 + 6x - (a2 + 2a – 8) = 0

Solution

We write, 6x = (a + 4)x – (a – 2)x as

x2 × [-(a2 + 2a – 8)] = -(a2 + 2a – 8)x2

= (a + 4)x × [-(a – 2)x]

∴ x2 + 6x – (a2 + 2a – 8) = 0

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

⇒ x[x + (a + 4)] – (a – 2)[x + (a = 4)] = 0

⇒ [x + (a + 4)][x – (a – 2)] = 0

⇒ x + (a + 4) = 0 or x – (a – 2) = 0

⇒ x = - (a + 4) = 0 x = a – 2

Hence, -(a + 4) and (a – 2) are the roots of the given equation.

46. abx2 + (b2 – ac)x – bc = 0

Solution

abx2 + (b2 – ac)x – bc = 0

⇒ abx2 + b2x – acx – bc = 0

⇒ bx(ax + b) – c(ax + b) = 0

⇒ (bx – c) (ax + b) = 0

⇒ bx – c = 0 or ax + b = 0

⇒ x = c/b or x = -b/a

Hence, the roots of the equation are c/b and -b/a.

47. x2 – 4ax – b2 + 4a2 = 0

Solution

We write, -4ax = -(b + 2a)x + (b – 2a)x as

x2 × (-b2 + 4a2) = (-b2 + 4a2)x2

= -(b + 2a)x × (b – 2a)x

∴ x2 – 4ax – b2 + 4a2 = 0

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

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

⇒ [x – (b + 2a)][x + (b – 2a)] = 0

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

⇒ x = 2a + b or x = -(b – 2a)

⇒ x = 2a + b or x = 2a – b

Hence, (2a + b) and (2a – b) are the roots of the given equation.

48. 4x2 – 2(a2 + b2)x + a2b2 = 0

Solution

4x2 – 2(a2 + b2)x + a2b2 = 0

⇒ 4x2 – 2a2x – 2b2x + a2b2 = 0

⇒ 2x(2x – a2) – b2(2x – a2) = 0

⇒ (2x – b2)(2x – a2) = 0

⇒ 2x – b2 = 0 or 2x – a2 = 0

⇒ x = b2/2 or x = a2/a

Hence, the roots of equation are b2/2 and a2/2.

49. 12abx2– (9a2 – 8b2)x – 6ab = 0

Solution

Given:

12abx2 – (9a2 – 8b2)x – 6ab = 0

⇒ 12abx2 – 9a2x + 8b2x – 6ab = 0

⇒ 3ax(4bx – 3a) + 2b(4bx – 3a) = 0

⇒ (3ax + 2b)(4bx - 3a) = 0

⇒ 3ax + 2b = 0 or 4bx – 3a = 0

⇒ x = -2b/3a or x = 3a/4b

Hence, the roots of the equation are -2b/3a and 3a/4b.

50. a2b2x2+ b2x – a2x – 1 = 0.

Solution

Given:

a2b2x + b2x – a2x – 1 = 0

⇒ b2x(a2x + 1) – 1(a2x + 1) = 0

⇒ (b2x – 1)(a2x + 1) = 0

⇒ (b2x – 1) = 0 or (a2x + 1) = 0

⇒ x = 1/b2 and -1/a2 are the roots of the given equation.

51. 9x2– 9(a + b)x + (2a2+ 5ab + 2b2) = 0

Solution

We write, -9(a + b)x = -3(2a + b)x – 3(a + 2b)x as

9x2 × (2a2 + 5ab + 2b2)

= 9(2a+ 5ab + 2b2)x2

= [-3(2a + b)x] × [-3(a + 2b)x]

∴ 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0

⇒ 9x2 – 3(2a + b)x – 3(a + 2b)x + (2a + b)(a + 2b) = 0

⇒ 3x[3x – (2a + b)] – (a + 2b)[3x – (2a + b)] = 0

⇒ [3x – (2a + b)][3x – (a + 2b)] = 0

⇒ 3x – (2a + b) = 0 or 3x – (a + 2b) = 0

⇒ x = (2a + b)/3 or x = (a + 2b)/3

Hence, (2a + b)/3 and (a + 2b)/3 are the roots of the given equation.

52. 16/x – 1 = 15/(x + 1), x ≠ 0 , -1

Solution

16/x – 1 = 15/(x + 1), x 4≠ 0, 1

⇒ 16/x – 1 = 15/(x + 1) = 1

⇒ (16x + 16 – 15x)/x(x + 1) = 1

⇒ (x + 16)/(x2 + x) = 1

⇒ x2 + x = x + 16 (Cross multiplication)

⇒ x2 – 16 = 0

⇒ (x + 4)(x – 4) = 0

⇒ x + 4 = 0 or x – 4 =0

⇒ x = -4 or x = 4

Hence, -4 and 4 are the roots of the given equation.

53. 4/x – 3 = 5/(2x + 3), x ≠ 0, -(3/2)

Solution

4/x – 3 = 5/(2x + 3), x ≠ 0, -(3/2)

⇒ 4/x – 5/(2x + 3) = 3

⇒ (8x + 12 – 5x)/x(2x + 3) = 3

⇒ (3x + 12)/(2x2 + 3x) = 3

⇒ (x + 4)/(2x2 + 3x) = 1

⇒ 2x2 + 3x = x + 4 (Cross multiplication)

⇒ 2x2 + 2x – 4 = 0

⇒ x2 + x – 2 = 0

⇒ x2 + 2x – x – 2 = 0

⇒ x(x + 2) – 1(x + 2) = 0

⇒ (x + 2)(x – 1) = 0

⇒ x + 2 = 0 or x – 1 = 0

⇒ x = -2 or x = 1

Hence, -2 and 1 are the roots of the given equation.

54. 3/(x + 1) - 1/2 = 2/(3x - 1), x = -1, 1/3

Solution

3/(x + 1) - 1/2 = 2/(3x - 1), x ≠ -1, 1/3

⇒ 3/(x + 1) – 2/(3x – 1) = 1/2

⇒ (9x – 3 – 2x – 2)/(x + 1)(3x – 1) = ½

⇒ (7x – 5)/(3x2 + 2x – 1) = 1/2

⇒ 3x2 + 2x – 1 = 14x – 10 (Cross multiplication)

⇒ 3x2 – 12x + 9 = 0

⇒ x2 – 4x + 3 = 0

⇒ x2 – 3x – x + 3 = 0

⇒ x(x – 3) – 1(x – 3) = 0

⇒ (x – 3)(x – 1) = 0

⇒ x – 3 = 0 or x – 1 = 0

⇒ x = 3 or x = 1

Hence, 1 and 3 are the roots of the given equation.

55. 1/(x – 1) – 1/(x + 5) = 6/7, x ≠ 1, -5

Solution

1/(x – 1) – 1/(x + 5) = 6/7, x ≠ 1, -5

⇒ (x + 5 – x + 1)/(x – 1)(x + 5) = 6/7

⇒ 6/(x2 + 4x – 5) = 6/7

⇒ x2 + 4x – 5 = 7

⇒ x2 + 4x – 12 = 0

⇒ x2 + 6x – 2x – 12 = 0

⇒ x(x + 6) – 2(x + 6) = 0

⇒ (x + 6)(x – 2) = 0

⇒ x + 6 = 0 or x – 2 = 0

⇒ x = -6 or x = 2

Hence, -6 and 2 are the roots of the given equation.

56. 1/(2a + b + 2x) = 1/2a + 1/b + 1/2x

Solution

1/(2a + b + 2x) = 1/2a + 1/b + 1/2x

⇒ 1/2a + b + 2x) – 1/2x = 1/2a + 1/b

⇒ (2x – 2a – b – 2x)/(2x(2a + b + 2x) = (2a + b)/2ab

⇒ -(2a + b)/(4x2 + 4ax + 2bx) = (2a + b)/2ab

⇒ 4x2 + 4ax + 2bx = -2ab

⇒ 4x2 + 4ax + 2bx = -2ab

⇒ 4x2 + 4ax + 2bx + 2ab = 0

⇒ 4x(x + a) + 2b(x + a) = 0

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

⇒ x + a(4x + 2b) = 0

⇒ x + a = 0 or 4x + 2b = 0

⇒ x = -a or x = -(b/2)

Hence, -a and -(b/2) are the roots of the given equation.

57. (x + 3)/(x – 2) - (1 – x)/x = 4.1/4 , x ≠ 2, 0

Solution

Given:

(x + 3)/(x – 2) – (1 – x)/x = 17/4

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

⇒ [x2 + 3x – (x – 2 – x2 + 2x)]/(x2 – 2x) = 17/4

⇒ (x2 + 3x + x2 – 3x + 2)/(x2 – 2x) = 17/4

⇒ (2x2 + 2)/(x2 – 2x) = 17/4

⇒ 8x2 + 8 = 17x2 – 34x

⇒ -9x2 + 34x – 8 = 0

⇒ 9x2 – 34x – 8 = 0

⇒ 9x2 – 36x +2x – 8 = 0

⇒ 9x(x – 4) + 2(x – 4) = 0

⇒ (x – 4)(9x + 2) = 0

⇒ x – 4 = 0 or 9x + 2 = 0 ≠

⇒ x = 4 or x = -2/9

Hence, the roots of the equation are 4 and -2/9.

58. (3x – 4)/7 + 7/(3x – 4) = 5/2, x ≠ 4/3

Solution

(3x – 4)/7 + 7/(3x – 4) = 5/2, x ≠ 4/3

⇒ (3x – 4)2 + 49/7(3x – 4) = 5/2

⇒ 9x2 – 24x + 16 + 49)/(21x – 28) = 5/2

⇒ 18x2 – 48x + 130 = 105x – 140

⇒ 18x2 – 153x + 270 = 0

⇒ 2x2 – 17x + 30 = 0

⇒ 2x2 – 12x – 5x + 30 = 0

⇒ 2x(x – 6) – 5(x – 6) = 0

⇒ (x – 6)(2x – 5) = 0

⇒ x – 6 = 0 or 2x – 5 = 0

⇒ x = 6 or x = 5/2

Hence, 6 and 5/2 are the roots of the given equation.

59. x/(x – 1) + (x – 1)/x = 4.1/4, x ≠ 0, 1

Solution

x/(x – 1) + (x – 1)/x = 4.1/4, x ≠ 0, 1

⇒ x2  + (x – 1)2/(x(x – 1)) = 17/4

⇒ (x2 + x2 – 2x + 1)/(x3 – x) = 17/4

⇒ (2x2 – 2x + 1)/(x2 – 1) = 17/4

⇒ 8x2 – 8x + 4 = 17x2 – 17x

⇒ 9x2 – 9x – 4 = 0

⇒ 9x2 – 12x + 3x – 4 = 0

⇒ 3x(3x – 4) + 1(3x – 4) = 0

⇒ (3x – 4)(3x + 1) = 0

⇒ (3x – 4)(3x + 1) = 0

⇒ 3x – 4 = 0 or 3x + 1 = 0

⇒ x = 4/3 or x = -(1/3)

Hence, 4/3 and -1/3 are the roots of the given equation.

60. x/(x + 1) + (x + 1)/x = 2.4/15, x ≠ 0, -1

Solution

x/(x +1) + (x + 1)/x = 2.4/15, x ≠ 0, -1

⇒ x2 + (x + 1)2/x(x + 1) = 34/15

⇒ (x2 + x2 + 2x + 1)/(x2 + x) = 34/15

⇒ (2x2 + 2x + 1)/(x2 + 1) = 34/15

⇒ 30x2 + 30x + 15 = 34x2 + 34x

⇒ 4x2 + 4x – 15 = 0

⇒ 4x2 + 10x – 6x – 15x = 0

⇒ 2x(2x + 5) – 3(2x + 5) = 0

⇒ (2x + 5)(2x – 3) = 0

⇒ 2x + 5 = 0 or 2x – 3 = 0

⇒ x = -5/2 or 2x – 3 = 0

Hence, -5/2 and 3/2 are the roots of the given equation.

61. (x – 4)/(x – 5) + (x – 6)/(x – 7) = 3.1/3, x ≠ 5, 7

Solution

(x – 4)/(x – 5) + (x – 6)/(x – 7) = 3.1/3, x ≠5, 7

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

⇒ (x2 – 11x + 28 + x2 – 11x + 30)/(x2 – 12x + 35) = 10/3

⇒ (2x2 – 22x + 58)/(x2 – 12x + 35) = 10/3

⇒ (x2 – 11x + 29)/(x2 – 12x + 35) = 5/3

⇒ 3x2 – 33x + 87 = 5x2 – 60x + 175

⇒ 2x2 – 33x + 87 = 5x2 – 60x + 175

⇒ 2x2 – 27x + 88 = 0

⇒ 2x2 – 16x – 11x + 88 = 0

⇒ 2x(x – 8) – 11(x – 8) = 0

⇒ (x – 8)(2x – 11) = 0

⇒ x – 8 = 0 or 2x – 11 = 0

⇒ x = 8 or x = 11/2

Hence, 8 and 11/2 are the roots of the given equation.

62. (x – 1)/(x – 2) + (x – 3)/(x – 4) = 3.1/3, x ≠ 2, 4

Solution

(x – 1)/(x – 2) + (x – 3)/(x – 4) = 3.1/3, x ≠ 2, 4

⇒ (x – 1)(x – 4) + (x – 2)(x – 3)}/{(x – 2)(x – 4) = 10/3

⇒ (x2 – 5x + 4 + x2 – 5x + 6)/(x2 – 6x + 8) = 10/3

⇒ (2x2 – 10x + 10)/(x2 – 6x + 8) = 10/3

⇒ (x2 – 5x + 5)/(x2 – 6x + 8) = 5/3

⇒ 3x2 – 15x + 15 = 5x2 – 30x + 40

⇒ 2x2 – 15x + 25 = 0

⇒ 2x2 – 10x – 5x + 25 = 0

⇒ 2x(x – 5) – 5(x – 5) = 0

⇒ (x – 5)(2x – 5) = 0

⇒ x – 5 = 0 or 2x – 5 = 0

⇒ x = 5 or x = 5/2

Hence, 5 and 5/2 are the roots of the given equation.

63. 1/(x – 2) + 2/(x – 1) = 6/x, x ≠ 0, 1, 2

Solution

1/(x – 2) + 2/(x – 1) = 6/x

⇒ (x – 1) + 2(x – 2)/(x – 1)(x – 2) = 6/x

⇒ (3x – 5)/(x2 – 3x + 2) = 6/x

⇒ 3x2 – 5x = 6x2 – 18x + 12 [On cross multiplying]

⇒ 3x2 – 13x + 12 = 0

⇒ 3x2 – (9 + 4)x + 12 = 0

⇒ 3x2 – 9x – 4x + 12 = 0

⇒ 3x(x – 3) – 4(x – 3) = 0

⇒ (3x – 4)(x – 3) = 0

⇒ 3x – 4 = 0 or x – 3 = 0

⇒ x = 4/3 or x = 3

64. 1/(x + 1) + 2/(x + 2) = 5/(x + 4), x ≠ -1, -2, -4

Solution

1/(x + 1) + 2/(x + 2) = 5/(x + 4), x ≠ -1, -2, -4

⇒ (x + 2 + 2x + 2)/(x + 1)(x + 2) = 5/(x + 4)

⇒ (3x + 4)/(x2 + 3x + 2) = 5/(x + 4)

⇒ (3x + 4)(x + 4) = 5(x2 + 3x + 2)

⇒ 3x2 + 16x + 16 = 5x2 + 15x + 10

⇒ 2x2 – x – 6 = 0

⇒ 2x2 – 4x + 3x – 6 = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ 3x2 + 16x + 16 = 5x2 + 15x + 10

⇒ 2x2 – x – 6 = 0

⇒ 2x2 – 4x + 3x – 6 = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ x – 2 = 0 or 2x + 3 = 0

⇒ x = 2 or x = -3/2

Hence, 2 and -3/2 are the roots of the given equation.

65. 3(3x – 1)/(2x + 3) – 2(2x + 3)/(3x – 1) = 5, x ≠ 1/3, -3/2

Solution

3(3x – 1)/(2x + 3) – 2(2x + 3)/(3x – 1) = 5, x ≠ 1/3, -3/2

⇒ 3(3x – 1)2 – 2(2x + 3)2/(2x + 3)(3x – 1) = 5

⇒ 3(9x2 – 6x + 1) – 2(4x2 + 12x + 9)/(6x2 + 7x – 3) = 5

⇒ (27x2 – 18x + 3 – 8x2 – 24x – 18)/(6x2 + 7x – 3) = 5

⇒ (19x2 – 42x – 15)/(6x2 + 7x – 3) = 5

⇒ 19x2 – 42x – 15 = 30x2 + 35x – 15

⇒ 11x2 + 77x = 0

⇒ 11x(x + 7) = 0

⇒ x = 0 or x + 7 = 0

⇒ x = 0 or x = - 7

Hence, 0 and -7 are the roots of the given equation.

66. 3(7x + 1)/(5x – 3) – 4(5x – 3)/(7x + 1) = 11, x ≠ 3/5, - 1/7

Solution

3(7x + 1)/(5x – 3) – 4(5x – 3)/(7x + 1) = 11, x ≠ 3/5, - 1/7

⇒ {3(7x + 1)2 – 4(5x – 3)2}/(5x – 3)(7x + 1) = 11

⇒ 3(49x2 +1 4x + 1) – 4(25x3 – 30x + 9)/(35x2 – 16x – 3) = 11

⇒ (147x2 + 42x + 3 – 100x2 + 120x – 36)/(35x2 – 16x – 3) = 11

⇒ (47x2 + 162x – 33)/(35x2 – 16x – 3) = 11

⇒ 338x(x – 1) = 0

⇒ x = 0 or x – 1 = 0

⇒ x = 0 x – 1 = 0

⇒ x = 0 or x = 1

Hence, 0 and 1 are the roots of the given equation.

67. (4x – 3)/(2x + 1) – 10(2x + 1)/(4x – 3) = 3, x ≠ -1/2, 3/4

Solution

Given:

(4x – 3)/(2x – 1) – 10(2x + 1)/(4x – 3) = 3

Putting (4x – 3)/(2x + 1) = y, we get:

y - 10/y = 3

⇒ (y2 – 10)/y = 3

⇒ y2 – 10 = 3y [On cross multiplying]

⇒ y2 – 3y – 10 = 0

⇒ y2 – (5 – 2)y – 10 = 0

⇒ y2 – 5y + 2y – 10 = 0

⇒ y(y – 5) + 2(y – 5) = 0

⇒ (y – 5)(y + 2) = 0

⇒ y – 5 = 0 or y + 2 = 0

⇒ y = 5 or y = -2

Case I:

If y = 5, we get:

⇒ (4x – 3)/(2x + 1) = 5

⇒ 4x – 3 = 5(2x + 1) [On cross multiplying]

⇒ 4x – 3 = 10x + 5

⇒ -6x = 8

⇒ -6x = 8

⇒ x = 8/6

⇒ x = -(4/3)

Case II:

If y = -2, we get:

(4x – 3)/(2x + 1) = -2

⇒ 4x – 3 = -2(2x + 1)

⇒ 4x – 3 = -4x – 2

⇒ 8x = 1

⇒ x = 1/8

Hence, the roots of the equation are –(4/3) and 1/8.

68. (x/(x + 1))2 – 5(x/(x + 1)) + 6 = 0, x ≠ b, a

Solution

(x/(x + 1))2 – 5(x/(x + 1)) + 6 = 0

Putting x/(x + 1) = y, we get:

y2 – 5y + 6 = 0

⇒ y2 – 5y + 6 = 0

⇒ y2 – (3 + 2) y + 6 = 0

⇒ y2 – 3y – 2y + 6 = 0

⇒ y(y – 3) – 2(y – 3) = 0

⇒ (y – 3)(y – 2) = 0

⇒ y – 3 = 0 or y – 2 = 0

⇒ y = 3 or y = 2

Case I:

If y = 3, we get

x/(x + 1) = 3

⇒ x = 3(x + 1) [On cross multiplying]

⇒ x = 3x + 3

⇒ x = -3/2

Case II:

If y = 2, we get:

x(x + 1) = 2

⇒ x = 2(x + 1)

⇒ x = 2x + 2

⇒ - x = 2

⇒ x = -2

Hence, the roots of the equation are -3/2 and -2.

69. a/(x – b) + b(x – a) = 2, x ≠ b, a

Solution

a/(x – b) + b/(x – a) = 2

⇒ (a – x + b)[2x – (a + b)] = 0

⇒ a – x + b = 0 or 2x – (a + b) = 0

⇒ x = a + b or x = (a + b)/2

Hence, the roots of the equation are (a + b) and (a + b)/2.

70. a/(ax – 1) + b/(bx – 1) = (a + b), x ≠ 1/a, 1/b

Solution

a/(ax – 1) + b/(bx – 1) = (a + b)

⇒ [a/(ax – 1) – b] + [b/(bx – 1) – a] = 0

⇒ [a – b(ax – 1)/(ax – 1)] + [b – a(bx – 1)/(bx – 1)] = 0

⇒ (a – abx + b)/(ax – 1) + (a – abx + b)/(bx – 1) = 0

⇒ (a – abx + b)[1/(ax – 1) + 1/(bx – 1)] = 0

⇒ (a – abx + b)[(bx – 1) + (ax – 1)/(ax – 1)(bx – 1)] = 0

⇒ (a – abx + b)[(a + b)x – 2/(ax – 1)(bx – 1)] = 0

⇒ (a – abx + b)[(a + b)x – 2] = 0

⇒ a – abx + b = 0 or (a + b)x – 2 = 0

⇒ x = (a + b)/ab or x = 2/(a + b)

Hence, the roots of the equation are (a + b)/ab and 2/(a + b).

71. 3(x + 2) + 3-x = 10

Solution

3(x + 2) + 3-x = 10

3x.9 + 1/3x = 10

Let 3x be equal to y.

∴ 9y + 1/y = 10

⇒ 9y2 + 1 = 10y

⇒ 9y2 – 10y + 1 = 0

⇒ (y – 1)(9y – 1) = 0

⇒ y – 1 = 0 or 9y – 1 = 0

⇒ y = 1 or y = 1/9

⇒ 3xx = 1 or 3x = 1/9

⇒ 3x = 30 or 3x = 3-2

⇒ x = 0 or x = -2

Hence, 0 and -2 are the roots of the given equation.

72. 4(x +1) + 4(1 – x) = 10

Solution

Given:

4(x + 1) + 4(1 – x) = 10

⇒ 4x .4 + 41.1/44 = 10

Let 4x be y.

∴ 4y + 4/y = 10

⇒ 4y2 – 10y + 4 = 0

⇒ 4y2 – 8y – 2y + 4 = 0

⇒ 4y(y – 2) – 2(y – 2) = 0

⇒ y = 2 or y = 2/4 = 1/2

⇒ 4x = 2 or 1/2

⇒ 4x = 22x or 21 or 22x = 2-1

⇒ x = ½ or x = -1/2

Hence, 1/2 and -1/2 are roots of the given equation.

73. 22x – 3.2(x + 2) + 32 = 0

Solution

22x – 3.2(x + 2) + 32 = 0

⇒ (2x)2 – 3.2x.22 + 32 = 0

Let 2x be y.

∴ y2 – 12y + 32 = 0

⇒ y2 – 8y – 4y + 32 = 0

⇒ y(y – 8) – 4(y – 8) = 0

⇒ (y – 8) = 0 or (y – 4) = 0

⇒ y = 8 or y = 4

∴ 2x = 8 or 2x = 4

⇒ 2x = 23 or 2x = 22

⇒ x – 2 or 3

Hence, 2 and 3 are the roots of the given equation.