# How do you simplify (sqrta-sqrt5)^2?

May 17, 2018

See a solution process below:

#### Explanation:

The rule for this special form of quadratic equation is:

${\left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right)}^{2} = \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) = {\textcolor{red}{x}}^{2} - 2 \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{y}}^{2}$

Substitute:

$\textcolor{red}{\sqrt{a}}$ for $\textcolor{red}{x}$

$\textcolor{b l u e}{\sqrt{5}}$ for $\textcolor{b l u e}{y}$

Giving:

${\left(\textcolor{red}{\sqrt{a}} - \textcolor{b l u e}{\sqrt{5}}\right)}^{2} \implies$

$\left(\textcolor{red}{\sqrt{a}} - \textcolor{b l u e}{\sqrt{5}}\right) \left(\textcolor{red}{\sqrt{a}} - \textcolor{b l u e}{\sqrt{5}}\right) \implies$

${\left(\textcolor{red}{\sqrt{a}}\right)}^{2} - 2 \textcolor{red}{\sqrt{a}} \textcolor{b l u e}{\sqrt{5}} + {\left(\textcolor{b l u e}{\sqrt{5}}\right)}^{2} \implies$

$\textcolor{red}{a} - 2 \sqrt{\textcolor{red}{a} \textcolor{b l u e}{5}} + \textcolor{b l u e}{5}$