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

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Answer 1 · 2017-06-06T14:48:44

$r=6$

$sqrt(r+3)=r-3$

$r+3=(r-3)^2$

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

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

$(r-1)(r-6)=0$

$r=6$ or $r=1$

$sqrt(6+3)-=6-3$ so $r=6$ is a valid solution

$sqrt(1+3)!=1-3$ so $r=1$ is not a valid solution

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