Question

giải phương trình: 2x²+4x=\(\sqrt{x^2+2x-4}+14\)

Answer 1

\(2x^2+4x=\sqrt{x^2+2x-4}+14\)

\(\Leftrightarrow2\left(x^2+2x-4\right)-\sqrt{x^2+2x-4}-6=0\)

Đặt \(\sqrt{x^2+2x-4}=a\left(a\ge0\right)\)

\(\Leftrightarrow2a^2-a-6=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-2=0\\2a+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=2\left(N\right)\\a=-\frac{3}{2}\left(L\right)\end{matrix}\right.\)

Với \(a=2\) :

\(\Leftrightarrow\sqrt{x^2+2x-4}=2\)

\(\Leftrightarrow x^2+2x-4=4\)

\(\Leftrightarrow x^2+2x-8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)