Problem of the Week

Problem 20 Solution

Congratulations to Yat Sum Ho from Cambourne Village College for successfully solving this problem!

i)
Assume that $\sqrt{3+2\sqrt{2}}$ can be written in the form of $\sqrt{a^2+2ab+b^2}$ , which will allow us to write it as $\sqrt{(a+b)^2}$ = $a+b$ .

If this is true, $a^2 + b^2$ = 3 and $2ab$ = $2\sqrt{2}$ $\implies ab=\sqrt{2}$

Having $a = 1$ and $b = \sqrt{2}$ satisfy these two equations, which yields

$\sqrt{3+2\sqrt{2}}$ = $1+\sqrt{2}$

ii) $\sqrt{4\sqrt{2} + 4\sqrt{3-2\sqrt{2}}}$

To begin with, we can factor out a 2 from this square root, giving us $2\sqrt{\sqrt{2} + \sqrt{3-2\sqrt{2}}}$
The inner square root looks quite similar to part (i), so it is reasonable to try approach it with the same strategy.