Question

5 and 7) Radical equations

Answer 1

5)    $\bf \sqrt{3r+36}=\sqrt{-9-2r}\impliedby \textit{squaring both sides} \\\\\\ (\sqrt{3r+36})^2=(\sqrt{-9-2r})^2\implies 3r+36=-9-2r$

solve for "r"

6)
$\bf a=\sqrt{-24+10a}\impliedby \textit{squaring both sides} \\\\\\ (a)^2=(\sqrt{-24+10a})^2\implies a^2=-24+10a \\\\\\ a^2\quad \underline{-10a}\quad \underline{+24}=0\qquad now\qquad \begin{cases} -6-4\to \underline{-10} \\\\ -6\times -4=\underline{+24} \end{cases}\qquad thus \\\\\\ (a-6)(a-4)=0\implies \begin{cases} a-6=0\implies &a=6\\\\ a-4=0\implies &a=4 \end{cases}$