# How do you solve sqrt(x-2)=x-2?

Jul 19, 2017

$x = 2$ or $x = 3$

#### Explanation:

$\sqrt{x - 2} = x - 2$

If we let $\alpha = x - 2$, then $\sqrt{\alpha} = \alpha$. We know that this can only be the case if $\alpha = 0$ or $\alpha = 1$. Then $x = 0 + 2$ or $x = 1 + 2$.

Jul 19, 2017

The solutions are $S = \left\{2 , 3\right\}$

#### Explanation:

Squaring the LHS and the RHS

$\sqrt{x - 2} = x - 2$

${\left(\sqrt{x - 2}\right)}^{2} = {\left(x - 2\right)}^{2}$

${x}^{2} - 4 x + 4 = \textcolor{red}{x - 2}$

${x}^{2} - 4 x \textcolor{red}{- x} + 4 \textcolor{red}{+ 2} = 0$

${x}^{2} - 5 x + 6 = 0$

Factorising

$\left(x - 2\right) \left(x - 3\right) = 0$

Therefore,

$\left(x - 2\right) = 0$, $\implies$, $x = 2$

$\left(x - 3\right) = 0$, $\implies$, $x = 3$

Verification

$\sqrt{2 - 2} = 2 - 2$

$\sqrt{3 - 2} = 3 - 2$