Finding the square root of $6-4\sqrt{2}$

Question

I found this standupmaths video on YouTube about the A4 paper puzzle.

I really liked the puzzle and managed to get the answer by using a calculator. However, the answer (which I won't spoil), led me to think that the equation to solve it might simplify - which it does.

In the middle of the simplification, I got this expression:

$$\sqrt{6-4\sqrt{2}}$$

which for other reasons I suspected to be equal to:

$$\ 2-\sqrt{2}$$

I tried squaring the above and, sure enough, it does give:

$$6-4\sqrt{2}$$

My question is, how would I have been able to find the square root of

$$6-4\sqrt{2}$$

if I hadn't been able to guess at it? Is there a standard technique? I've tried looking on the web but don't really even know what to search for.

Answer 1

One way of doing this would be to guess that $$\sqrt{6-4\sqrt{2}} = a+b\sqrt2$$ for some $a,b \in \mathbb Z$. Squaring both sides, we get $$6-4\sqrt2 = a^2+2b^2 + 2ab\sqrt2$$ which implies that $6=a^2+2b^2$ and $-4 = 2ab$. Since $a$ and $b$ are integers, there are two solutions: $(a,b) = (2,-1)$, and $(a,b) = (-2,1)$, and we reject the latter.

Answer 2

We have to assume that the nested radical can be rewritten as the sum of two other radicals (surds).

$\sqrt{6-4\sqrt{2}}=\sqrt{d}+\sqrt{e}$

Squaring both sides gives us $$6-4\sqrt{2}=d+e+2\sqrt{de}$$

This can be solved by finding $2$ numbers that sum to $6$ and multiply to $4^2\cdot 2/4=8$

($2\sqrt{de}=4\sqrt{2}\rightarrow de=8$)

Numbers $4$ and $2$ work, so $$\sqrt{6-4\sqrt{2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}$$

Answer 3

The numbers of the form $a+b\sqrt{2}$ share many arithmetic properties with the integers. That's why you might suspect an answer of that form to the question. So try one: $$ (a+b\sqrt{2})^2 = a^2 + 2b^2 + 2ab\sqrt{2} = 6-4\sqrt{2}. $$

Then it's easy to finish.

You can find out more on wikipedia (https://en.wikipedia.org/wiki/Quadratic_integer) but you'd have to know to search for "quadratic integer" to find that page.

Answer 4

set $$(a+b\sqrt{2})^2=6-4\sqrt{2}$$ and find the rational numbers $a$ and $b$

Answer 5

To expand on @Frank's answer, there is a general formula for simplifying expressions of the form $\sqrt{a + \sqrt{b}}$ by assuming the result will take the form $\sqrt{c}+\sqrt{d}$.

(If the original expression is of the form $\sqrt{a + q\sqrt{b}}$, recall that $q\sqrt{b}=\sqrt{q^2b}$. The formula we will derive will still work.)

$$\sqrt{a + \sqrt{b}}=\sqrt{c}+\sqrt{d}$$ $$\Rightarrow a + \sqrt{b}=c + d + 2\sqrt{c}\sqrt{d}$$ Assume $a=c+d$ and $b=4cd$ $$\Rightarrow c=a-d$$ $$\Rightarrow b=4(a-d)d$$ $$=4ad-4d^2$$ $$\Rightarrow 4d^2-4ad+b=0$$ Applying quadratic formula, $$d=\frac{4a\pm\sqrt{16a^2-4*4*b}}{8}$$ $$=\frac{4a\pm\sqrt{16a^2-16b}}{8}$$ $$=\frac{a\pm\sqrt{a^2-b}}{2}$$ The two solutions to this equation (presuming they exist) will be $c$ and $d$, respectively. This will only be a simplifying tool if the radicand, $a^2-b$, is a perfect square.

In the case of the OP's question, $a=6$ and $b=32 \Rightarrow a^2-b=36-32=4$, so the formula works. $c$ and $d$ are therefore equal to $$\frac{6\pm{2}}{2}=2,4$$ $$\Rightarrow \sqrt{6 + 4\sqrt{2}}=\sqrt{2}+\sqrt{4}$$ $$=2 + \sqrt{2}$$