# Question #6766a

Oct 13, 2016

#### Answer:

Show that

${\left(4 \sqrt{5} - 3 \sqrt{2}\right)}^{2} = 98 - \sqrt{10}$

#### Explanation:

Using that ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$, and $\sqrt{a} \sqrt{b} = \sqrt{a b}$ for $a , b \ge 0$, we have

${\left(4 \sqrt{5} - 3 \sqrt{2}\right)}^{2} = {\left(4 \sqrt{5}\right)}^{2} - 2 \left(4 \sqrt{5}\right) \left(3 \sqrt{2}\right) + {\left(3 \sqrt{2}\right)}^{2}$

$= {4}^{2} {\left(\sqrt{5}\right)}^{2} - 24 \sqrt{5} \sqrt{2} + {3}^{2} {\left(\sqrt{2}\right)}^{2}$

$= 16 \left(5\right) - 24 \sqrt{5 \cdot 2} + 9 \cdot 2$

$= 98 - 24 \sqrt{10}$