Question

Answer 1

\(\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{7}{3}\end{matrix}\right.\)

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