# How do you solve sqrt(r+3)=r-3 and check your solution?

Jun 6, 2017

$r = 6$

#### Explanation:

$\sqrt{r + 3} = r - 3$

$r + 3 = {\left(r - 3\right)}^{2}$

$r + 3 = {r}^{2} - 6 r + 9$

${r}^{2} - 7 r + 6 = 0$

$\left(r - 1\right) \left(r - 6\right) = 0$

$r = 6$ or $r = 1$

$\sqrt{6 + 3} \equiv 6 - 3$ so $r = 6$ is a valid solution

$\sqrt{1 + 3} \ne 1 - 3$ so $r = 1$ is not a valid solution