# How do you evaluate (2\sqrt { 2} - \sqrt { 7} ) ^ { 2}?

Dec 7, 2016

${\left(2 \sqrt{2} - \sqrt{7}\right)}^{2}$ simplifies to $- 4 \sqrt{14} + 15$.

#### Explanation:

For the square of any binomial, it follows this general rule: ${\left(x - y\right)}^{2} = {x}^{2} - 2 x y + {y}^{2}$. We can apply this here where $x$ is $2 \sqrt{2}$ and $y$ is $\sqrt{7}$

${\left(2 \sqrt{2} - \sqrt{7}\right)}^{2}$
$= {\left(2 \sqrt{2}\right)}^{2} - 2 \left(2 \sqrt{2}\right) \left(\sqrt{7}\right) + {\left(\sqrt{7}\right)}^{2}$
$= 8 - 4 \sqrt{7 \cdot 2} + 7$
$= - 4 \sqrt{14} + 15$

So, our final simplified answer is $- 4 \sqrt{14} + 15$.