# Find the roots of the following equations:

(i) x - 1/x = 3, x ≠ 0

(ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, 7

**Solution:**

Convert the given equation in the form of ax^{2} + bx + c = 0 and by using the quadratic formula, find the roots.

(i) x - 1/x = 3, x ≠ 0

x - 1/x = 3, x ≠ 0 can be rewritten as (multiplying both sides by x):

x^{2} - 1 = 3x

x^{2} - 3x - 1 = 0

Comparing this against the standard form of quadratic equation that is ax^{2} + bx + c = 0, we find that a = 1, b = - 3, c = -1

b^{2} - 4ac = (-3)^{2} - 4(1)(-1)

= 9 + 4

= 13 > 0

Therefore, x = (-b ± √b^{2} - 4ac) / 2a

x = (3 ± √13) / 2

The roots are (3 + √13) / 2, (3 - √13) / 2

(ii) 1/(x + 4) - 1/(x - 7) = 11/ 30, x ≠ - 4, 7

By cross multiplying we get:

[(x - 7) - (x + 4)] / (x + 4)(x - 7) = 11/ 30

[x - 7 - x - 4] / x^{2} + 4x - 7x - 28 = 11/ 30

(- 11) / (x^{2} - 3x - 28) = 11/30

- 11 × 30 = 11(x^{2} - 3x - 28)

- 30 = (x^{2} - 3x - 28) [Cancelling 11 from both sides of the equation ]

x^{2} - 3x - 28 + 30 = 0

x^{2} - 3x + 2 = 0

Comparing this against the standard form ax^{2} - bx + c = 0 , we find that a = 1, b = - 3 c = 2

b^{2} - 4ac = (-3)^{2} - 4(1)(2)

= 9 - 8

= 1 > 0

Therefore, real roots exist for this quadratic equation.

x = (3 + √9 - 8) / 2, (3 - √9 - 8) / 2

x = (3 + 1)/2 and x = (3 - 1)/2

x = 4 / 2 and x = 2 / 2

x = 2 and x = 1

Therefore, roots are 2, 1.

**Video Solution:**

## Find the roots of the following equations: (i) x - 1/x = 3, x ≠ 0 (ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, 7

### Class 10 Maths NCERT Solutions - Chapter 4 Exercise 4.3 Question 3:

Find the roots of the following equations: (i) x - 1/x = 3, x ≠ 0 (ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, 7

The roots of the first quadratic equation is x = 3 + √13 / 2 or x = 3 - √13 / 2 and for second quadratic equation the roots are 2 and 1