Question

Find the root of the following equation.

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

Use quadratic formula to solve:

`1/(x+4) - 1/(x-7) = 11/30`

Answer 1

`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`

⇒ (x + 4)(x − 7) = −30

⇒ x² − 3x − 28 = 30

⇒ x² − 3x + 2 = 0

⇒ x² − 2x − x + 2 = 0

⇒ x(x − 2) − 1(x − 2) = 0

⇒ (x − 2)(x − 1) = 0

⇒ x = 1 or 2

Answer 2

`1/(x+4) - 1/(x-7) = 11/30`

⇒ `((x−7)−(x+4))/((x−7)−(x+4)) = 11/30`

⇒ `(x-7-x-4)/((x+4)(x-7)) = 11/30`

⇒ `(-11)/((x+4)(x-7)) = 11/30`

⇒ −11 × 30 = 11(x + 4)(x − 7)

⇒ `(−11 × 30)/11` = x² − 7x + 4x − 28

⇒ −30 = x² − 3x − 28

⇒ x² − 3x + 2 = 0

Now we have,

x² − 3x + 2 = 0

Comparing x² − 3x + 2 = 0 with ax² + bx + c = 0 we get,

a = 1, b = −3 and c = 2.

We know that,

x = `(−b ± sqrt(b^2−4ac))/(2a)`

Substituting values of a, b and c in above equation we get,

x = `(- (-3) ± sqrt((-3)^2−4(1)(2)))/(2(1))`

= `(3 +- sqrt(9 - 8))/2`

= `(3 +- sqrt1)/2`

= `(3 +- 1)/2`

= `(3 + 1)/2` or `(3 - 1)/2`

= `4/2 or 2/2`

= 2 or 1