Question

If m and n are roots of the equation `1/x - 1/(x - 2) = 3`; where x ≠ 0 and x ≠ 2; find m × n.

Answer 1

Given quadratic equation is `1/x - 1/(x - 2) = 3`

⇒ x – 2 – x = 3x(x – 2)

⇒ –2 = 3x² – 6x

⇒ 3x² – 6x + 2 = 0

a = 3, b = – 6, c = 2 

D = b² – 4ac

= (– 6)² – 4(3)(2)

= 36 – 24

= 12

⇒ `x = (-b ± sqrt(b^2 - 4ac))/(2a)`

⇒ `x = (-6 ± sqrt12)/(2 xx 3)`

⇒ `x=(6 ± sqrt(2xx2xx3))/6`

⇒ `x = (6 ± (2sqrt3))/6`

⇒ `x = (6 + 2sqrt3)/6` or `x = (6 - 2sqrt3)/6`

Since, m and n are roots of the equation, we have

⇒ `m = (6 + 2sqrt3)/6  n = (6 - 2sqrt3)/6`

⇒ `m xx n = (6 + 2sqrt3)/6 xx(6 - 2sqrt3)/6`

= `((6)^2 - (2sqrt3)^2)/36`

= `(36 - 4xx3)/36` 

= `(36 - 12)/36`

= `24/36`

= `2/3`