can anyone help me with this inequalities??

$\displaystyle \sqrt{x-2}\leq x-4$

this is what i've done...

$\displaystyle (\sqrt{x-2})^2 \leq (x-4)^2$

$\displaystyle x-2\leq x^2-8x+16$

so....

$\displaystyle x\leq 3$or $\displaystyle x\geq6$

and

roots rule

$\displaystyle x-2\geq0$

$\displaystyle x\geq2$

so the area will be at

$\displaystyle 2\leqx\leq3$ or $\displaystyle x\geq6$

but,

when i've try with x=2.5

then i'll be $\displaystyle \sqrt{0.5} \leq -1.5$ (it's absolutely wrong)

why????

regards,

pencil09