Question

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

Answer 1

Consider the given inequality.

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

So,

$|x-3|\ge 2x^2-10x+12$

Now, for $x\le 3$, we have

$3-x\ge 2x^2-10x+12$

$2x^2-9x+9\le 0$

$x\in [\frac{3}{2},3]$

Again, for $x\ge 3$, we have

$x-3\ge 2x^2-10x+12$

$2x^2-11x+15\le 0$

$x\in [\frac{5}{2},3]$

This doesn’t satisfy $x\ge 3$. So, we will ignore this value. Also, since, $x=3$ is a root of the equation $x^2-5x+6=0$, it will make the denominator zero. So, this value will also be ignored. Therefore, the required set of values is,

$\Rightarrow x\in [\frac{3}{2},3)$

Hence, this is the required result.