Answer
The student didn't check the square roots for additional solutions. The solutions of the equation are ±1 and ±5.
Work Step by Step
$r= \sqrt {-6r-5}$
$r^2= (\sqrt {-6r-5})^2$
$r^2 = -6r-5$
$r^2+6r+5 = -6r-5+6r+5$
$r^2+6r+5 = 0$
$(r+5)(r+1)=0$
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 $\sqrt 1 = -1$
