# How do you simplify ( sqrt 2 + sqrt 5)^4?

Apr 11, 2016

${\left(\sqrt{2} + \sqrt{5}\right)}^{4} = 89 + 28 \sqrt{10}$

#### Explanation:

Note that ${\left(\sqrt{2} + \sqrt{5}\right)}^{4} = {\left({\left(\sqrt{2} + \sqrt{5}\right)}^{2}\right)}^{2}$

So let us square $\left(\sqrt{2} + \sqrt{5}\right)$ twice:

${\left(\sqrt{2} + \sqrt{5}\right)}^{2}$

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

$= 2 + 2 \sqrt{10} + 5$

$= 7 + 2 \sqrt{10}$

Then:

${\left(7 + 2 \sqrt{10}\right)}^{2}$

$= {7}^{2} + 2 \left(7\right) \left(2 \sqrt{10}\right) + {\left(2 \sqrt{10}\right)}^{2}$

$= 49 + 28 \sqrt{10} + 40$

$= 89 + 28 \sqrt{10}$

Check

Let's check the calculation a different way.

From the Binomial Theorem:

${\left(a + b\right)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$

So:

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

$= {\left(\sqrt{2}\right)}^{4} + 4 {\left(\sqrt{2}\right)}^{3} \left(\sqrt{5}\right) + 6 {\left(\sqrt{2}\right)}^{2} {\left(\sqrt{5}\right)}^{2} + 4 \left(\sqrt{2}\right) {\left(\sqrt{5}\right)}^{3} + {\left(\sqrt{5}\right)}^{4}$

$= 4 + 8 \sqrt{10} + 60 + 20 \sqrt{10} + 25$

$= 89 + 28 \sqrt{10}$