# How can you simplify sqrt(28-5sqrt(12)) ?

Dec 12, 2016

#### Answer:

$\sqrt{28 - 5 \sqrt{12}} = 5 - \sqrt{3}$

#### Explanation:

For a start:

$\sqrt{12} = \sqrt{{2}^{2} \cdot 3} = 2 \sqrt{3}$

So:

$\sqrt{28 - 5 \sqrt{12}} = \sqrt{28 - 10 \sqrt{3}}$

Can this be simplified further?

Let us attempt to find rational $a , b$ such that:

$28 - 10 \sqrt{3} = {\left(a + b \sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{28 - 10 \sqrt{3}} = \left({a}^{2} + 3 {b}^{2}\right) + 2 a b \sqrt{3}$

Equating coefficients:

$\left\{\begin{matrix}{a}^{2} + 3 {b}^{2} = 28 \\ 2 a b = - 10\end{matrix}\right.$

From the second equation, we find:

$b = - \frac{5}{a}$

Substituting $- \frac{5}{a}$ for $b$ in the first equation, we get:

$28 = {a}^{2} + \frac{75}{a} ^ 2$

Subtracting $28$ from both sides and multiplying by ${a}^{2}$ we get:

$0 = {\left({a}^{2}\right)}^{2} - 28 \left({a}^{2}\right) + 75$

$\textcolor{w h i t e}{0} = \left({a}^{2} - 25\right) \left({a}^{2} - 3\right) = \left(a - 5\right) \left(a + 5\right) \left(a - \sqrt{3}\right) \left(a + \sqrt{3}\right)$

Since we want $a$ to be rational, we have:

$a = \pm 5$

If $a = 5$ then $b = - \frac{5}{a} = - 1$, resulting in $5 - \sqrt{3}$ which is positive.

If $a = - 5$ then $b = - \frac{5}{a} = 1$, resulting in $- 5 + \sqrt{3}$ which is negative.

Since we want the non-negative square root, we want $5 - \sqrt{3}$