# A curious square root (Part 1)

Here’s a square root that looks like a total mess:

$\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}}$

Just look at this monstrosity, which has a triply-embedded square root! But then look what happens when I plug into a calculator:

Hmmm. How is that possible?!?!

I’ll give the answer after the thought bubble, if you’d like to think about it before seeing the answer.

Let’s start from the premise that $\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}} = 3$ and work backwards. This isn’t the best of logic — since we’re assuming the thing that we’re trying to prove in the first place — but it’s a helpful exercise to see exactly how this happened.

$\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}} = 3$

$5 - \sqrt{6} + \sqrt{22+8\sqrt{6}} = 9$

$\sqrt{22+8\sqrt{6}} = 4 + \sqrt{6}$

$22 + 8 \sqrt{6} = (4 + \sqrt{6})^2$

This last line is correct, using the formula $(a+b)^2 = a^2 + 2ab + b^2$. So, running the logic from bottom to top (and keeping in mind that all of the terms are positive), we obtain the top equation.

This suggests a general method for constructing such hairy square roots. To begin, start with any similar expression, such as

$(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3$

$(2 - \sqrt{3})^2 = 7 - 4\sqrt{3}$

Then we create a nested square root:

$2 - \sqrt{3} = \sqrt{7 - 4\sqrt{3}}$

Then I get rid of the square root on the left hand side:

$2 = \sqrt{3} + \sqrt{7 - 4\sqrt{3}}$

Then I add or subtract something so that the left-hand side becomes a perfect square.

$25 = 23 + \sqrt{3} + \sqrt{7 - 4 \sqrt{3}}$

Finally, I take the square root of both sides:

$5 = \sqrt{23 + \sqrt{3} + \sqrt{7 - 4 \sqrt{3}}}$

Then I show the right-hand side to my students, ask them to plug into their calculators, and ask them to figure out what happened.