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 = 2x^2 - 15x + 25`

⇒ `2x^2 - 15x - 3x + 25 + 8` = 0

⇒ `2x^2 - 18x + 33` = 0

On company with `ax^2 + bx + c` = 0, we get

a = 2, b = – 18 and c = 33

∴ Discriminant, D = `b^2 - 4ac`

= `(-18)^2 - 4 xx 2(33)`

= `324 - 264`

= 60 > 0

Therefore, the equation `2x^2 - 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`