Simplify $\dfrac{2\left(\sqrt2 + \sqrt6\right)}{3\sqrt{2+\sqrt3}}$
The answer to this question is $\frac{4}{3}$ in a workbook.
How would I simplify $\sqrt{2+\sqrt3}$ $?$ If it was something like $\sqrt{3 + 2\sqrt2}$ , I would have simplified it as follows:
$\sqrt{3 + 2\sqrt2}$
$=$ $\sqrt{(\sqrt2)^2 + 2(\sqrt2)(1) + (1)^2}$
$=$
$\sqrt{(\sqrt2 + 1)^2}$
$=$
$\sqrt2 + 1$
But I can't simplify $\sqrt{2+\sqrt3}$ like that as $2+\sqrt3$ is can't be written as squares of two numbers. Is there any other method?
Best Answer
Note that$$\left(\frac{2\left(\sqrt2+\sqrt6\right)}{3\sqrt{2+\sqrt3}}\right)^2=\frac{4\left(8+4\sqrt3\right)}{9\left(2+\sqrt3\right)}=\frac{16}9=\left(\frac43\right)^2.$$