## Precalculus (6th Edition) Blitzer

The provided statement is false and the correct statement is “The solution set of ${{x}^{2}}>25$ is $\left( -\infty ,-5 \right)\cup \left( 5,\infty \right).$ ”
The provided inequality is: ${{x}^{2}}>25$ Rewrite, the provided inequality as: ${{x}^{2}}-25>0$ In order to get the solution of ${{x}^{2}}>25$ , first equate the function to $0$ , that is solve ${{x}^{2}}-25=0$. Then, \begin{align} & {{x}^{2}}-25=0 \\ & {{x}^{2}}-{{5}^{2}}=0 \\ & \left( x-5 \right)\left( x+5 \right)=0 \end{align} Equating each factor to 0, $x=-5$ or $x=5$ Therefore, the solution set of ${{x}^{2}}>25$ is $\left( -\infty ,-5 \right)\cup \left( 5,\infty \right).$ And thereby, the provided statement is false. The provided statement is false and the correct statement is “The solution set of ${{x}^{2}}>25$ is $\left( -\infty ,-5 \right)\cup \left( 5,\infty \right).$ ”