## Precalculus (6th Edition) Blitzer

The solution of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ is $\left\{ 1,5 \right\}$.
Consider the quadratic equation, $0=-2{{\left( x-3 \right)}^{2}}+8$ Add the term $2{{\left( x-3 \right)}^{2}}$ on both sides, \begin{align} & 0+2{{\left( x-3 \right)}^{2}}=-2{{\left( x-3 \right)}^{2}}+8+2{{\left( x-3 \right)}^{2}} \\ & 2{{\left( x-3 \right)}^{2}}=8 \end{align} Divide both sides by 2, \begin{align} & 0+2{{\left( x-3 \right)}^{2}}=-2{{\left( x-3 \right)}^{2}}+8+2{{\left( x-3 \right)}^{2}} \\ & \frac{2}{2}{{\left( x-3 \right)}^{2}}=\frac{8}{2} \\ & {{\left( x-3 \right)}^{2}}=4 \end{align} Apply the square root property, \begin{align} & x-3=\pm \sqrt{4} \\ & x-3=\pm 2 \\ & x=3\pm 2 \end{align} First take the positive sign, \begin{align} & x=3+2 \\ & =5 \end{align} Secondly, take the negative sign, \begin{align} & x=3-2 \\ & =1 \end{align} Thus, the two values of x are 5 and 1. Thus, the solutions of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ are 1 and 5. Check it by substituting the value of x in the equations. Check: For, $x=1$ \begin{align} & 0\overset{?}{\mathop{=}}\,-2{{\left( 1-3 \right)}^{2}}+8 \\ & 0\overset{?}{\mathop{=}}\,-2{{\left( -2 \right)}^{2}}+8 \\ & 0\overset{?}{\mathop{=}}\,-8+8 \\ & 0\overset{?}{\mathop{=}}\,0 \end{align} Which is true. For, $x=5$ \begin{align} & 0\overset{?}{\mathop{=}}\,-2{{\left( 5-3 \right)}^{2}}+8 \\ & 0\overset{?}{\mathop{=}}\,-2{{\left( 2 \right)}^{2}}+8 \\ & 0\overset{?}{\mathop{=}}\,-8+8 \\ & 0\overset{?}{\mathop{=}}\,0 \end{align} Which is true. Thus, the solution of the quadratic equation $0=-2{{\left( x-3 \right)}^{2}}+8$ is $\left\{ 1,5 \right\}$.