Question

5 and 7) Radical equations

Answer 1

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

solve for "r"

6)

`\bf a=√(-24+10a)\impliedby \textit{squaring both sides} \\\\\\ (a)^2=(√(-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* -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}`