# Simplify sqrt(29-2sqrt(28)) ?

Feb 26, 2017

$1 + \sqrt{7}$

#### Explanation:

$\sqrt{29 - 2 \sqrt{28}} = \sqrt{29 - 2 \sqrt{{2}^{2} \cdot 7}}$

$\textcolor{w h i t e}{\sqrt{29 - 2 \sqrt{28}}} = \sqrt{29 - 4 \sqrt{7}}$

Is there a number of the form $a + b \sqrt{7}$ with square $29 - 4 \sqrt{7}$?

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

Equating coefficients, we want to find $a , b$ satisfying:

$\left\{\begin{matrix}{a}^{2} + 7 {b}^{2} = 29 \\ 2 a b = - 4\end{matrix}\right.$

In particular, we would like $29 - 7 {b}^{2}$ to be a perfect square.

We quickly find that $b = \pm 2$ works, resulting in:

${a}^{2} = 29 - 7 {b}^{2} = 29 - 7 \left({2}^{2}\right) = 1$

So $a = \pm 1$

Then from $2 a b = - 4$ we find two possible solution pairs:

$\left(a , b\right) = \left(1 , - 2\right) \text{ }$ or $\text{ } \left(a , b\right) = \left(- 1 , 2\right)$

The second of these results in the positive square root:

$\sqrt{29 - 4 \sqrt{7}} = - 1 + 2 \sqrt{7}$

Next we have:

$9 + \sqrt{29 - 4 \sqrt{7}} = 9 + \left(- 1 + 2 \sqrt{7}\right) = 8 + 2 \sqrt{7}$

and we would like to find the square root of this.

Attempt to solve:

$8 + 2 \sqrt{7} = {\left(c + \mathrm{ds} q r t \left(7\right)\right)}^{2} = \left({c}^{2} + 7 {d}^{2}\right) + 2 c \mathrm{ds} q r t \left(7\right)$

Hence:

$\left\{\begin{matrix}{c}^{2} + 7 {d}^{2} = 8 \\ 2 c d = 2\end{matrix}\right.$

We can fairly quickly spot that $\left(c , d\right) = \left(1 , 1\right)$ results in a positive square root:

$\sqrt{8 + 2 \sqrt{7}} = 1 + \sqrt{7}$