## Algebra 1

$r= \sqrt {-6r-5}$ $r^2= (\sqrt {-6r-5})^2$ $r^2 = -6r-5$ $r^2+6r+5 = -6r-5+6r+5$ $r^2+6r+5 = 0$ $(r+5)(r+1)=0$ The student said that both -1 and -5 are solutions. $-5 =\sqrt {-6*-5-5}$ $-5 = \sqrt{30-5}$ $-5 = \sqrt{25}$ $\sqrt{25} = 5$ and $\sqrt{25} = -5$ $-1 = \sqrt {-6*-1-5}$ $-1 = \sqrt{6-5}$ $\sqrt 1 = 1$ and $\sqrt 1 = -1$