## Precalculus (6th Edition) Blitzer

The solution set of equation $\sqrt{x-3}+5=x$ is $\left\{ \left. 7 \right\} \right.$.
Consider the provided equation, $\sqrt{x-3}+5=x$ The provided equation can be written as: \begin{align} & \sqrt{x-3}+5=x \\ & \sqrt{x-3}=x-5 \end{align} Square both sides $x-3={{\left( x-5 \right)}^{2}}$ Apply ${{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy$ \begin{align} & x-3={{x}^{2}}+25-10x \\ & {{x}^{2}}+25-10x-x+3=0 \\ & {{x}^{2}}-11x+28=0 \\ \end{align} Apply quadratic formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Here a is 1, b is $-11$ and c is $28$ \begin{align} & x=\frac{11\pm \sqrt{{{\left( -11 \right)}^{2}}-4\left( 1 \right)\left( 28 \right)}}{2\left( 1 \right)} \\ & =\frac{11\pm \sqrt{121-112}}{2} \\ & =\frac{11\pm 3}{2} \end{align} On further simplification \begin{align} & x=\frac{11+3}{2} \\ & =\frac{14}{2} \\ & =7 \end{align} Or \begin{align} & x=\frac{11-3}{2} \\ & =\frac{8}{2} \\ & =4 \end{align} Check: For $x=7$ Put $x=7$ in $\sqrt{x-3}+5=x$ \begin{align} & \sqrt{7-3}+5\overset{?}{\mathop{=}}\,7 \\ & \sqrt{4}+5\overset{?}{\mathop{=}}\,7 \\ & 2+5\overset{?}{\mathop{=}}\,7 \\ & 7\overset{?}{\mathop{=}}\,7 \end{align} Which is true. For $x=4$ Put $x=4$ in $\sqrt{x-3}+5=x$ \begin{align} & \sqrt{4-3}+5\overset{?}{\mathop{=}}\,4 \\ & \sqrt{1}+5\overset{?}{\mathop{=}}\,4 \\ & 1+5\overset{?}{\mathop{=}}\,4 \\ & 6\overset{?}{\mathop{=}}\,4 \end{align} Which is not true. Therefore, the solution set of the equation $\sqrt{x-3}+5=x$ is $\left\{ \left. 7 \right\} \right.$.