Question

Solving radical equations

check for extraneous solutions

`√(2x + 30) = x + 3`

Answer 1

`\bf √(2x+30)=x+3\impliedby \textit{we'll squares both sides} \\\\\\ (√(2x+30))^2=(x+3)^2\implies 2x+30=x^2+6x+9 \\\\\\ 0=x^2+4x-21\implies 0=(x+7)(x-3)\implies x= \begin{cases} -7\\ 3 \end{cases}`

I don't see an extraneous value there, -7 checks out well if you plug it in the original equation, and so does 3.

Answer 2

I don’t know this question