[Math] How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$

algebra-precalculusfractions

I was tackling through an olympiad practice book when I saw one of these problems:

If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ – \ 1 \over\sqrt{x}}$?

The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$

Best Answer

Hint $\ \ $ Squaring it, we get $\ \rm\dfrac{40+16\sqrt{6}}{5+2\sqrt{6}}\, =\, 8$

Remark $\ \ $ Generally $\rm\ \ \dfrac{ n\!-\!1 + \sqrt{n^2\!-\!1}}{ \sqrt{ n\ +\ \sqrt{n^2\!-\!1}}}\, =\, \sqrt{2(n\!-\!1)}\ $ for $\rm\: n\ge 1,\:$ with analogous proof.

For another example note $\, \dfrac{\quad\ 3 + \sqrt{11}}{\sqrt{10+ \sqrt{99}}}\, =\, \sqrt{2},\ $ by $\rm\:n = 10\:$ above.

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