# I'm stuck trying to figure out 8 square of 2 and 13 square root 3 in radical form i know the decimal is 33.83036899 but don't now the radical form can someone please help?

## 8 $\sqrt{2}$ +13 $\sqrt{3}$

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Feb 9, 2018

$8 \sqrt{2} + 13 \sqrt{3}$ is already in simplest form.

#### Explanation:

The square roots $\sqrt{2}$ and $\sqrt{3}$ are independent over the rationals.

In fact the set of numbers of the form $a + b \sqrt{2} + c \sqrt{3} + \mathrm{ds} q r t \left(6\right)$ where $a , b , c , d$ are rational is closed under addition, subtraction, multiplication and division of non-zero numbers, with the coefficients $a , b , c , d$ being unique.

For example, $\sqrt{3}$ is not expressible in the form $a + b \sqrt{2}$ for rational $a , b$.

Attempting to solve:

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

We find that we require $a = 0$ and/or $b = 0$ in order that the right hand side is rational.

If $b = 0$ then we have ${a}^{2} = 3$, which is not possible.

If $a = 0$ then we have $2 {b}^{2} = 3$, so ${\left(2 b\right)}^{2} = 6$, which is not possible either since $\sqrt{6}$ is irrational.

Having shown $\sqrt{3}$ is not expressible in the form $a + b \sqrt{2}$, suppose:

$a + b \sqrt{2} + c \sqrt{3} + \mathrm{ds} q r t \left(6\right) = 0$

Then if at least one of $a , b , c , d$ is non-zero, then ${c}^{2} - 2 {d}^{2} \ne 0$ and we find:

$\sqrt{3} = - \frac{a + b \sqrt{2}}{c + \mathrm{ds} q r t \left(2\right)}$

$\textcolor{w h i t e}{\sqrt{3}} = - \frac{\left(a + b \sqrt{2}\right) \left(c - \mathrm{ds} q r t \left(2\right)\right)}{{c}^{2} - 2 {d}^{2}}$

$\textcolor{w h i t e}{\sqrt{3}} = \frac{a d - b c}{{c}^{2} - 2 {d}^{2}} + \frac{b c - a d}{{c}^{2} - 2 {d}^{2}} \sqrt{2}$

which contrdicts our earleir finding.

Therefore:

$a + b \sqrt{2} + c \sqrt{3} + \mathrm{ds} q r t \left(6\right) = 0 \text{ " => " } a = b = c = d = 0$

Hence the representation $a + b \sqrt{2} + c \sqrt{3} + \mathrm{ds} q r t \left(6\right)$ is unique.

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