Question

Find whether the following equation have real roots. If real roots exist, find them.

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

Answer 1

Given equation is `1/(2x - 3) + 1/(x - 5) = 1, x ≠ 3/2, 5`

⇒ `(x - 5 + 2x - 3)/((2x - 5)(x - 5))` = 1

⇒ `(3x - 8)/(2x^2 - 5x - 10x + 25)` = 1

⇒ `(3x - 8)/(2x^2 - 15x + 25)` = 1

⇒ 3x – 8 = 2x² – 15x + 25

⇒ 2x² – 15x – 3x + 25 + 8 = 0

⇒ 2x² – 18x + 33 = 0

On company with ax² + bx + c = 0, we get

a = 2, b = – 18 and c = 33

∴ Discriminant, D = b² – 4ac

= (–18)² – 4 × 2(33)

= 324 – 264

= 60 > 0

Therefore, the equation 2x² – 18x + 33 = 0 has two distinct real roots.

Roots, `x = (-b +- sqrt(D))/(2a)`

= `(-(-18) +- sqrt(60))/(2(2))`

= `(18 +- 2sqrt(15))/4`

= `(9 +- sqrt(15))/2`

= `(9 + sqrt(15))/2, (9 - sqrt(15))/2`