Phương trình $\sqrt {2{x^2} + 6x + 1}  = x + 2$ có nghiệm là:

Ta có $\sqrt {2{x^2} + 6x + 1}  = x + 2$$\Leftrightarrow \left\{ \begin{array}{l}x + 2 \ge 0\\2{x^2} + 6x + 1 = {\left( {x + 2} \right)^2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\{x^2} - 2x - 3 = 0\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\{x^2} - 3x + x - 3 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\x\left( {x - 3} \right) + \left( {x - 3} \right) = 0\end{array} \right.$$\Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\\left( {x + 1} \right)\left( {x - 3} \right) = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge  - 2\\\left[ \begin{array}{l}x =  - 1\\x = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - 1\\x = 3\end{array} \right.$

Vậy phương trình có nghiệm $x =  - 1;x = 3.$