# How do you solve sqrt x = x - 6?

From the relation $\sqrt{x} = x - 6$ we have that $x \ge 0$ and

$x - 6 \ge 0 \implies x \ge 6$.Finally solutions must be in $\left[6 , + \infty\right)$

hence

$\sqrt{x} = x - 6$

${\left(\sqrt{x}\right)}^{2} = {\left(x - 6\right)}^{2}$

$x = {\left(x - 6\right)}^{2}$

$x = {x}^{2} - 12 x + 36$

${x}^{2} - 12 x - x + 36 = 0$

${x}^{2} - 13 x + 36 = 0$

$\left(x - 4\right) \cdot \left(x - 9\right) = 0$

${x}_{1} = 4$ and ${x}_{2} = 9$

The only acceptable solution is $x = 9$