## Precalculus (6th Edition) Blitzer

The solution set of equation $2{{x}^{2}}-3x-2=0$ is $\left\{ \left. -\frac{1}{2},2 \right\} \right.$.
Consider the provided equation: $2{{x}^{2}}-3x-2=0$ On factorizing, we get: \begin{align} & 2{{x}^{2}}-4x+x-2=0 \\ & 2x\left( x-2 \right)+1\left( x-2 \right)=0 \\ & \left( x-2 \right)\left( 2x+1 \right)=0 \end{align} Now use the principle of zero products: $\left( x-2 \right)=0$ or $\left( 2x+1 \right)=0$ From here \begin{align} & \left( x-2 \right)=0 \\ & x=2 \end{align} Or \begin{align} & \left( 2x+1 \right)=0 \\ & 2x=-1 \\ & x=\frac{-1}{2} \end{align} Check: For $x=2$ Put $x=2$ in $2{{x}^{2}}-3x-2=0$ \begin{align} & 2{{\left( 2 \right)}^{2}}-3\left( 2 \right)-2\overset{?}{\mathop{=}}\,0 \\ & 2\left( 4 \right)-6-2\overset{?}{\mathop{=}}\,0 \\ & 8-8\overset{?}{\mathop{=}}\,0 \\ & 0=0 \end{align} For $x=\frac{-1}{2}$ Put $x=\frac{-1}{2}$ in $2{{x}^{2}}-3x-2=0$ \begin{align} & 2{{\left( \frac{-1}{2} \right)}^{2}}-3\left( \frac{-1}{2} \right)-2\overset{?}{\mathop{=}}\,0 \\ & 2\left( \frac{1}{4} \right)+\frac{3}{2}-2\overset{?}{\mathop{=}}\,0 \\ & \frac{1}{2}-\frac{1}{2}\overset{?}{\mathop{=}}\,0 \\ & 0=0 \end{align} Which is true. Therefore, the solution set of the equation $2{{x}^{2}}-3x-2=0$ is $\left\{ \left. -\frac{1}{2},2 \right\} \right.$.