## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

${{x}^{2}}-6x-5=0$
$3-\sqrt{14}\ \text{and }3+\sqrt{14}$ It can be expressed as, $x=3+\sqrt{14}$ and $x=3-\sqrt{14}$ \begin{align} & x-\left( 3+\sqrt{14} \right)=0 \\ & \left( x-3 \right)-\sqrt{14}=0 \end{align} Or, \begin{align} & x-\left( 3-\sqrt{14} \right)=0 \\ & \left( x-3 \right)+\sqrt{14}=0 \\ \end{align} Apply the zero-product property $\left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0$ \begin{align} & \left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0 \\ & \left( x-3 \right)\left( x-3 \right)+\sqrt{14}\left( x-3 \right)-\sqrt{14}\left( x-3 \right)-{{\left( \sqrt{14} \right)}^{2}}=0 \end{align} Combine like terms: Here ${{i}^{2}}=1$ which is a complex entry. ${{\left( x-3 \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ \begin{align} & {{x}^{2}}+9-6x-\left( 14 \right)=0 \\ & {{x}^{2}}-6x+9-14=0 \\ & {{x}^{2}}-6x-5=0 \end{align}