How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$

Question

This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expression.

Answer 1

Such roots can be computed by a Simple Denesting Rule:. $ $ The radicand is $\, 8-\sqrt{48} = 4(2-\sqrt3)$

Here $\ 8-\sqrt{48}\ $ has norm $= 16.\:$ $\rm\ \color{blue}{Subtracting\ out}\,\ \sqrt{norm}\ = 4\,\ $ yields $\,\ 4-\sqrt{48} $

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{8},\,\ \ \rm so\ \ \ \color{brown}{Dividing\ it\ out}\,\ $ of the above yields $\ \dfrac{\sqrt{16}-\sqrt{48}}{\sqrt{8}} =\!\!\!\underbrace{\sqrt{2}-\sqrt 6}_{\rm negate\ for\ root > 0}$

Answer 2

Suppose $8-4\sqrt 3= (a-b\sqrt 3)^2$

This gives $a^2+3b^2=8$ and $2ab=4$ so that $ab=2$

This means that $a^2b^2+3b^4=8b^2$ or $4+3b^4=8b^2$ or $$3b^4-8b^2+4=0=(3b^2-2)(b^2-2)$$

So either $b^2=2$ or $b^2=\frac 23$.

We have also $a^2b^2=4$ so that $a^2=2$ or $a^2=6$

This should suffice to identify the solution which corresponds to the principal value (non-negative real for square root of non-negative real) of the square root.

Since the two square roots in the original are both positive, we have a unique solution. But the alternate possible signs indicate that a quartic is in the background. The other solutions of the quartic correspond to alternative choices of sign for the square root.

Answer 3

Note that \begin{align} 4(2-\sqrt{3}) = 8 - 4\sqrt{3} = (\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2 = (\sqrt{6}-\sqrt{2})^2. \end{align} Therefore, $$ \sqrt{6} -\sqrt{2} = \sqrt{4(2-\sqrt{3})}. $$