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I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution

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

My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea how to transform the calculation to symbolically get to that result.

(I can factor out one $\sqrt{2}$ from both terms, but that does not lead me anywhere, either)

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    $\begingroup$ Hint: $6-4\sqrt{2}=\left(2-\sqrt{2}\right)^2$. $\endgroup$
    – user236182
    Commented Jan 15, 2016 at 18:47
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    $\begingroup$ ok thanks, that helps me prove it. But how do you come to this solution (or do you just know because you have seen it a lot of times)? $\endgroup$
    – mihi
    Commented Jan 15, 2016 at 18:48
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    $\begingroup$ Simplifying such expressions is called Denesting Nested Radicals. It's easy once your calculator tells you it should equal $2$, because you're left with proving $\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$, which can be done with the hint I mentioned. $\endgroup$
    – user236182
    Commented Jan 15, 2016 at 18:51
  • $\begingroup$ Thanks (If my task would have been to prove the equation you wrote, I probably had the idea to square both sides and compare, too...) If you put that into an answer, I'll accept it :) $\endgroup$
    – mihi
    Commented Jan 15, 2016 at 18:55
  • $\begingroup$ You can square both sides, prove that the squared equality holds, then note that $2-\sqrt{2}>0$. This would finish the proof. The property used here is the following: If $a,b\ge 0$ and $a^2=b^2$, then $a=b$. $\endgroup$
    – user236182
    Commented Jan 15, 2016 at 19:55

4 Answers 4

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Start by trying to simplify $\sqrt{6-4\sqrt{2}}$. Let's assume there is some number $p+q\sqrt{2}$ for which $$\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$$ Squaring both sides gives $$6-4\sqrt{2} = (p + q\sqrt{2})^2 = p^2+2q^2 + 2pq\sqrt{2}$$ Comparing coefficients gives $6=p^2+2q^2$ and $-4=2pq$, i.e. $-2=pq$.

We need to solve $p^2+2q^2=6$ and $pq = -2$ simultaneously.

If $pq=-2$ then $q=-\frac{2}{p}$ and we can substitute this into $p^2+2q^2=6$. We get \begin{eqnarray*} p^2+2q^2 &=& 6 \\ \\ p^2 + 2\left(-\frac{2}{p}\right)^2 &=& 6 \\ \\ p^2 + \frac{8}{p^2} &=& 6 \\ \\ p^4+8 &=& 6p^2 \\ \\ p^4-6p^2+8 &=& 0 \\ \\ (p^2-2)(p^2-4) &=& 0 \end{eqnarray*}

Either $p^2-2=0$ or $p^2-4=0$, i.e. $p=\pm\sqrt{2}$ or $p=\pm 2$. The only valid solutions are $p = \pm2$ because we usually assume that $p$ and $q$ are rational numbers, i.e. fractions.

If $p=\pm 2$ then $pq=2$ gives $\pm 2q=-2$, and so $q=\mp 1$. Hence $$p+q\sqrt{2} = \pm(2-\sqrt{2})$$

Recall that $\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$ and since, by definition, $\sqrt{6-4\sqrt{2}} \ge 0$ we conclude $$\sqrt{6-4\sqrt{2}} = 2-\sqrt{2}$$

Finally: $$\sqrt{6-4\sqrt{2}} \ \ {\color{red}{+\sqrt{2}}}= 2-\sqrt{2} \ \ {\color{red}{+\sqrt{2}}} = 2$$

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Since \begin{align} 6-4\sqrt{2}&=4-2\cdot 2\cdot\sqrt{2}+2\\ &=(2-\sqrt{2})^2 \end{align} Where $2-\sqrt{2}>0$, it follows $\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$.

Then $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$

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Hint:

If the expression under the first radical is a perfect square, the double product $4\sqrt2$ factors as $2\cdot2\cdot\sqrt2$. Then you indeed have $6-4\sqrt2=2^2-2\cdot2\cdot\sqrt2+(\sqrt2)^2$.

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$$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{4^2}\sqrt{2}}+\sqrt{2}=$$ $$\sqrt{6-\sqrt{16}\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{16\cdot2}}+\sqrt{2}=$$ $$\sqrt{6-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{6^2}-\sqrt{32}}+\sqrt{2}=$$ $$\sqrt{\sqrt{36}-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{4\cdot9}-\sqrt{4\cdot8}}+\sqrt{2}=$$ $$\sqrt{2\sqrt{9}-2\sqrt{8}}+\sqrt{2}=\sqrt{2\left(\sqrt{9}-\sqrt{8}\right)}+\sqrt{2}=$$ $$\sqrt{2}\sqrt{\sqrt{9}-\sqrt{8}}+\sqrt{2}=\sqrt{2}\left(\sqrt{\sqrt{9}-\sqrt{8}}+1\right)=$$ $$\sqrt{2}\left(\sqrt{3-2\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2\left(\frac{3}{2}-\sqrt{2}\right)}+1\right)=$$ $$\sqrt{2}\left(\sqrt{2}\sqrt{\frac{3}{2}-\sqrt{2}}+1\right)= \sqrt{2}\left(\sqrt{2}\left(1-\frac{1}{\sqrt{2}}\right)+1\right)=$$ $$\sqrt{2}\left(\sqrt{2}-\frac{\sqrt{2}}{\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2}-1+1\right)=$$ $$\sqrt{2}\left(\sqrt{2}\right)=\left(\sqrt{2}\right)^2=2$$

EDIT:

$$\sqrt{\frac{3}{2}-\sqrt{2}}=\sqrt{\frac{3}{2}-\frac{2\sqrt{2}}{2}}=$$ $$\sqrt{\frac{3-2\sqrt{2}}{2}}=\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{4}}=$$ $$\frac{\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{4}}=\frac{2-\sqrt{2}}{2}=1-\frac{\sqrt{2}}{2}=1-\frac{1}{\sqrt{2}}$$

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  • $\begingroup$ Very detailed way of factoring out the $\sqrt{2}$. But I don't see how you get from $\sqrt{\frac32 - \sqrt2}$ to $1 - \frac1{\sqrt2}$. $\endgroup$
    – mihi
    Commented Jan 16, 2016 at 12:10
  • $\begingroup$ @mihi see my edit! $\endgroup$
    – Jan Eerland
    Commented Jan 16, 2016 at 14:01
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    $\begingroup$ which then boils down to having to know that $3-2\sqrt2$ is $\frac{(2-\sqrt2)^2}{2}$ - which is similar enough to knowing that $6-4\sqrt2 = (2-\sqrt2)^2$, which is the problem I stated... $\endgroup$
    – mihi
    Commented Jan 16, 2016 at 18:28

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