## Precalculus (6th Edition) Blitzer

The solution set of equation $x\left( x-2 \right)=4$ is $\left\{ \left. 1-\sqrt{5},1+\sqrt{5} \right\} \right.$.
Consider the equation: $x\left( x-2 \right)=4$ \begin{align} & x\left( x-2 \right)=4 \\ & {{x}^{2}}-2x=4 \\ & {{x}^{2}}-2x-4=0 \\ \end{align} The above equation is in the form of $a{{x}^{2}}+bx+c=0$ Apply quadratic formula. $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ Here a is 1, b is $-2$ and c is $-4$ \begin{align} & x=\frac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -4 \right)}}{2\left( 1 \right)} \\ & =\frac{2\pm \sqrt{4+16}}{2} \\ & =\frac{2\pm \sqrt{20}}{2} \\ & =\frac{2\pm \sqrt{5\cdot 4}}{2} \end{align} On further simplification, we get: $x=\frac{2\pm 2\sqrt{5}}{2}$ Take out the 2 common term: \begin{align} & x=\frac{2\left( 1\pm \sqrt{5} \right)}{2} \\ & =1\pm \sqrt{5} \end{align} Check: For $x=1+\sqrt{5}$ Put $x=1+\sqrt{5}$ in $x\left( x-2 \right)=4$ \begin{align} & \left( 1+\sqrt{5} \right)\left( 1+\sqrt{5}-2 \right)\overset{?}{\mathop{=}}\,4 \\ & \left( \sqrt{5}+1 \right)\left( \sqrt{5}-1 \right)\overset{?}{\mathop{=}}\,4 \\ & {{\left( \sqrt{5} \right)}^{2}}-{{1}^{2}}\overset{?}{\mathop{=}}\,4 \\ & 5-1\overset{?}{\mathop{=}}\,4 \end{align} Which is true. For $x=1-\sqrt{5}$ Put $x=1-\sqrt{5}$ in $x\left( x-2 \right)=4$ \begin{align} & \left( 1-\sqrt{5} \right)\left( 1-\sqrt{5}-2 \right)\overset{?}{\mathop{=}}\,4 \\ & \left( 1-\sqrt{5} \right)\left( -1-\sqrt{5} \right)\overset{?}{\mathop{=}}\,4 \\ & {{\left( -\sqrt{5} \right)}^{2}}-{{1}^{2}}\overset{?}{\mathop{=}}\,4 \\ & 5-1\overset{?}{\mathop{=}}\,4 \end{align} Which is true. Therefore, the solution set of the equation $x\left( x-2 \right)=4$ is $\left\{ \left. 1-\sqrt{5},1+\sqrt{5} \right\} \right.$.