# What is the rationalising factor of the given number?

## 2+sqrt(7+2sqrt(10)

Aug 13, 2018

$6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}$

#### Explanation:

I will assume that you are looking for a radical conjugate, that is an expression which when multiplied by the given expression gives a rational product.

Given:

$2 + \sqrt{7 + 2 \sqrt{10}}$

First we should check whether the expression $\sqrt{7 + 2 \sqrt{10}}$ can be simplified.

Note that in https://socratic.org/s/aTtiPKas I found that given:

$\sqrt{p + q \sqrt{r}} \text{ }$ with $p , q , r > 0$

then if ${p}^{2} - {q}^{2} r$ is a perfect square ${s}^{2}$ then:

$\sqrt{p + q \sqrt{r}} = \frac{\sqrt{2 p + 2 s}}{2} + \frac{\sqrt{2 p - 2 s}}{2}$

In our example, putting $p = 7$, $q = 2$, $r = 10$, then:

$s = \sqrt{{p}^{2} - {q}^{2} r} = \sqrt{{7}^{2} - {2}^{2} \left(10\right)} = \sqrt{49 - 40} = \sqrt{9} = 3$

So:

$\sqrt{7 + 2 \sqrt{10}} = \frac{\sqrt{2 \left(7\right) + 2 \left(3\right)}}{2} + \frac{\sqrt{2 \left(7\right) - 2 \left(3\right)}}{2}$

$\textcolor{w h i t e}{\sqrt{7 + 2 \sqrt{10}}} = \frac{\sqrt{20}}{2} + \frac{\sqrt{8}}{2}$

$\textcolor{w h i t e}{\sqrt{7 + 2 \sqrt{10}}} = \sqrt{5} + \sqrt{2}$

So:

$2 + \sqrt{7 + 2 \sqrt{10}} = 2 + \sqrt{5} + \sqrt{2}$

Since this involves two square roots, a suitable radical conjugate is formed by multiplying the variants of the expression $2 \pm \sqrt{5} \pm \sqrt{2}$ apart from the one we already have.

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

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

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

$= \left(2 + \sqrt{5} - \sqrt{2}\right) \left(7 - 4 \sqrt{5}\right)$

$= 7 \left(2 + \sqrt{5} - \sqrt{2}\right) - 4 \sqrt{5} \left(2 + \sqrt{5} - \sqrt{2}\right)$

$= \left(14 + 7 \sqrt{5} - 7 \sqrt{2}\right) - \left(8 \sqrt{5} + 20 - 4 \sqrt{10}\right)$

$= - 6 - \sqrt{5} - 7 \sqrt{2} + 4 \sqrt{10}$

Note that this is negative, so, let's negate it to get the slightly more attractive radical conjugate:

$6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}$

As a check, let's multiply this by $2 + \sqrt{5} + \sqrt{2}$ and see what we get:

$\left(2 + \sqrt{5} + \sqrt{2}\right) \left(6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}\right)$

$= 2 \left(6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}\right) + \sqrt{5} \left(6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}\right) + \sqrt{2} \left(6 + \sqrt{5} + 7 \sqrt{2} - 4 \sqrt{10}\right)$

$= \left(12 + 2 \sqrt{5} + 14 \sqrt{2} - 8 \sqrt{10}\right) + \left(6 \sqrt{5} + 5 + 7 \sqrt{10} - 20 \sqrt{2}\right) + \left(6 \sqrt{2} + \sqrt{10} + 14 - 8 \sqrt{5}\right)$

$= 31$

Aug 13, 2018

Here's another way to simplify...

#### Explanation:

One way of simplifying $\sqrt{7 + 2 \sqrt{10}}$ involves considering the quartic polynomial with integer coefficients of which it is one of the zeros.

The other zeros will be the variants:

$- \sqrt{7 + 2 \sqrt{10}}$, $\text{ } \sqrt{7 - 2 \sqrt{10}}$, $\text{ }$ and $\text{ } - \sqrt{7 - 2 \sqrt{10}}$

The quartic of which these are zeros can be expressed as:

${\left({x}^{2} - 7\right)}^{2} - 40$

$= {x}^{4} - 14 {x}^{2} + 9$

$= {x}^{4} - 6 {x}^{2} + 9 - 8 {x}^{2}$

$= {\left({x}^{2} - 3\right)}^{2} - {\left(2 \sqrt{2} x\right)}^{2}$

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

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

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

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

Hence zeros: $x = \pm \sqrt{2} \pm \sqrt{5}$

The greatest of these is $x = \sqrt{2} + \sqrt{5}$, which must be the greatest of $\pm \sqrt{7 \pm 2 \sqrt{10}}$, i.e. $\sqrt{7 + 2 \sqrt{10}}$

So:

$\sqrt{7 + 2 \sqrt{10}} = \sqrt{2} + \sqrt{5}$

From this point we can proceed as my other answer to multiply:

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