Question

Express the equation `(x - 2)/(x - 3) + (x - 4)/(x - 5) = 10/3`; (x ≠ 3, 5) as a quadratic equation in standard form. Hence, find the roots of the equation so formed.

Answer 1

`(x - 2)/(x - 3) + (x - 4)/(x - 5) = 10/3`

⇒ `((x - 2)(x - 5)(x - 4)(x - 3))/((x - 3)(x - 5)) = 10/3`

⇒ `(x^2 - 2x - 5x + 10 + x^2 - 4x - 3x + 12)/(x^2 - 5x - 3x + 15) = 10/3`

⇒ `(2x^2 − 14x + 22)/(x^2 − 8x + 15) = 10/3`

⇒ 6x² − 42x + 66 = 10x² − 80x + 150

⇒ 4x² − 38x + 84 = 0

⇒ 2x² − 19x + 42 = 0

a = 2, b = –19, с = 42

Applying the quadratic formula

x = `(-b ± sqrt(b^2 - 4ac))/(2a)`

x = `(-(-19) ± sqrt((-19)^2 - 4 xx 2 xx 42))/(2 xx 2)`

x = `(19 ± sqrt(361 - 336))/4`

x = `(19 ± sqrt25)/4`

x = `(19 ± 5)/4`

x = `(19 + 5)/4, (19 - 5)/4`

x = `24/4, 14/4`

Roots are 6, `7/2` and quadratic equation is 2x² − 19x + 42 = 0.