## Precalculus (6th Edition) Blitzer

The solution set of equation $\sqrt{8-2x}-x=0$ is $\left\{ \left. 2 \right\} \right.$.
Consider the provided equation, $\sqrt{8-2x}-x=0$ The above equation can be written as: \begin{align} & \sqrt{8-2x}-x=0 \\ & \sqrt{8-2x}=x \end{align} Square both sides \begin{align} & 8-2x={{x}^{2}} \\ & {{x}^{2}}+2x-8=0 \end{align} Apply quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Here a is 1, b is $2$ and c is $-8$ \begin{align} & x=\frac{-2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)} \\ & =\frac{-2\pm \sqrt{4+32}}{2} \\ & =\frac{-2\pm \sqrt{36}}{2} \\ & =\frac{-2\pm 6}{2} \end{align} On further simplification \begin{align} & x=\frac{-2+6}{2} \\ & =\frac{4}{2} \\ & =2 \end{align} Or \begin{align} & x=\frac{-2-6}{2} \\ & =\frac{-8}{2} \\ & =-4 \end{align} Check: For $x=2$ Put $x=2$ in $\sqrt{8-2x}-x=0$ \begin{align} & \sqrt{8-2\left( 2 \right)}-2\overset{?}{\mathop{=}}\,0 \\ & \sqrt{8-4}-2\overset{?}{\mathop{=}}\,0 \\ & \sqrt{4}-2\overset{?}{\mathop{=}}\,0 \\ & 2-2\overset{?}{\mathop{=}}\,0 \end{align} Which is true. For $x=-4$ Put $x=-4$ in $\sqrt{8-2x}-x=0$ \begin{align} & \sqrt{8-2\left( -4 \right)}-\left( -4 \right)\overset{?}{\mathop{=}}\,4 \\ & \sqrt{8+8}+4\overset{?}{\mathop{=}}\,4 \\ & \sqrt{16}+4\overset{?}{\mathop{=}}\,4 \\ & 4+4\overset{?}{\mathop{=}}\,4 \end{align} Therefore $8\overset{?}{\mathop{=}}\,4$ Which is not true. Therefore, the solution set of the equation $\sqrt{8-2x}-x=0$ is $\left\{ \left. 2 \right\} \right.$.