Question

Solve the following inequality:
$\frac{|x-3|}{x^2-5x+6}⩾2$

Answer 1

R.E.F image

$\frac{|x-3|}{x^2-5x+6}\ge 2$

Sol Case (i) $x\ge 3$

$\frac{x-3}{x^2-5x+6}\ge 6$

$\frac{x-3}{x^2-5x+6}-6\ge 0$

$\frac{x-3-6x^2+30x-36}{x^2-5x+6}\ge 0$

$\frac{-6x^2+37x-39}{x^2-5x+6}\ge 0$

$\frac{[-6x^2-18x-13x+39]}{x^2-5x+6}\ge 0$

$\frac{-6x(x-3)-13(x-3)}{x^2-5x+6}\ge 0$

$\frac{-(6x-13)(x-3)}{(x-5)(x-1)}\ge 0$

$-(x-3)(x-1)(x-5)(6x-13)\le 0$

$xϵ[3,5]$

$xϵ[1,\frac{11}{6}]$

So final Answer,

$xϵ[3,5]\cup [1,11/6]$