# How do you show that sqrt(2-sqrt3)/2=(sqrt6-sqrt2)/4?

Jun 8, 2017

First, note that any number in the form $a - \sqrt{b}$ when squared gives a number in the form $c - \sqrt{d}$.

In fact,

${\left(a - \sqrt{b}\right)}^{2} = \left({a}^{2} + b\right) - 2 a \sqrt{b} = \left({a}^{2} + b\right) - \sqrt{4 {a}^{2} b}$

The goal here is to reduce $\sqrt{2 - \sqrt{3}}$ by writing $2 - \sqrt{3}$ in the form ${\left(a - \sqrt{b}\right)}^{2}$, so the square root and the exponent of $2$ will cancel one another out.

We want:

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

Creating a system:

$\left\{\begin{matrix}2 = {a}^{2} + b \\ 4 {a}^{2} b = 3\end{matrix}\right.$

Yielding:

$\left\{\begin{matrix}{a}^{2} = 2 - b \\ {a}^{2} = \frac{3}{4 b}\end{matrix}\right.$

So:

$\frac{3}{4 b} = 2 - b$

$4 {b}^{2} - 8 b + 3 = 0$

$\left(2 b - 1\right) \left(2 b - 3\right) = 0$

$b = \frac{1}{2} , \frac{3}{2}$

And from ${a}^{2} = 2 - b$, these give:

$\left\{\begin{matrix}\left(a b\right) = \left(\sqrt{\frac{3}{2}} \frac{1}{2}\right) \\ \left(a b\right) = \left(\sqrt{\frac{1}{2}} \frac{3}{2}\right)\end{matrix}\right.$

Which gives the identity:

$2 - \sqrt{3} = {\left(\sqrt{\frac{3}{2}} - \sqrt{\frac{1}{2}}\right)}^{2}$

(You can verify this if you want)

And from here, we see that:

$\frac{\sqrt{2 - \sqrt{3}}}{2} = \frac{\sqrt{{\left(\sqrt{\frac{3}{2}} - \sqrt{\frac{1}{2}}\right)}^{2}}}{2} = \frac{\sqrt{\frac{3}{2}} - \sqrt{\frac{1}{2}}}{2}$

$= \frac{\frac{\sqrt{3} - 1}{\sqrt{2}}}{2} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

Multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$:

$= \frac{\sqrt{6} - \sqrt{2}}{4}$

Jun 8, 2017

Given: $\frac{\sqrt{2 - \sqrt{3}}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Multiply the left side by 1 in the form of $\frac{2}{2}$:

$\frac{2}{2} \frac{\sqrt{2 - \sqrt{3}}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

The 2 in the numerator goes inside the square root as ${2}^{2}$

$\frac{\sqrt{{2}^{2} \left(2 - \sqrt{3}\right)}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Distribute $4$:

$\frac{\sqrt{\left(8 - 4 \sqrt{3}\right)}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

The 4 goes inside the inner root as 16:

$\frac{\sqrt{\left(8 - \sqrt{48}\right)}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Break 8 into 6 + 2 and factor the root term:

$\frac{\sqrt{\left(6 - 2 \sqrt{6} \sqrt{2} + 2\right)}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Write the end terms as squares of roots:

$\frac{\sqrt{\left({\left(\sqrt{6}\right)}^{2} - 2 \sqrt{6} \sqrt{2} + {\left(\sqrt{2}\right)}^{2}\right)}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

It fits the pattern of a square of the difference:

$\frac{\sqrt{{\left(\sqrt{6} - \sqrt{2}\right)}^{2}}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Square root of a square

$\frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$