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

$16{{x}^{2}}-40x-8=0$
$\frac{5}{4}-\frac{\sqrt{33}}{4}\ \text{and }\frac{5}{4}+\frac{\sqrt{33}}{4}$ It can be expressed as: $x=\frac{5}{4}+\frac{\sqrt{33}}{4}$ and $x=\frac{5}{4}-\frac{\sqrt{33}}{4}$ \begin{align} & x-\left( \frac{5}{4}+\frac{\sqrt{33}}{4} \right)=0 \\ & \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4}=0 \end{align} Or, \begin{align} & x-\left( \frac{5}{4}-\frac{\sqrt{33}}{4} \right)=0 \\ & \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4}=0 \\ \end{align} Apply the zero-product property: $\left( \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4} \right)\left( \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4} \right)=0$ \begin{align} & \left( \left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4} \right)\left( \left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4} \right)=0 \\ & \left( x-\frac{5}{4} \right)\left( x-\frac{5}{4} \right)+\frac{\sqrt{33}}{4}\left( x-\frac{5}{4} \right)-\frac{\sqrt{33}}{4}\left( x-\frac{5}{4} \right)-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0 \end{align} Combine the like terms: ${{\left( x-\frac{5}{4} \right)}^{2}}-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ \begin{align} & {{\left( x-\frac{5}{4} \right)}^{2}}-{{\left( \frac{\sqrt{33}}{4} \right)}^{2}}=0 \\ & {{x}^{2}}+\frac{25}{16}-2x\cdot \frac{5}{4}-\frac{33}{16}=0 \\ & {{x}^{2}}-\frac{5x}{2}+\frac{25-33}{16}=0 \\ & {{x}^{2}}-\frac{5x}{2}-\frac{8}{16}=0 \end{align} Multiply the equation by $16$ to clear the fraction, \begin{align} & 16\left( {{x}^{2}}-\frac{5x}{2}-\frac{8}{16} \right)=0 \\ & 16{{x}^{2}}-40x-8=0 \end{align}