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How do you simplify: $$\sqrt{9-6\sqrt{2}}$$

A classmate of mine changed it to $$\sqrt{9-6\sqrt{2}}=\sqrt{a^2-2ab+b^2}$$ but I'm not sure how that helps or why it helps.

This questions probably too easy to be on the Math Stack Exchange but I'm not sure where else to post it.

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  • $\begingroup$ See How to simplify a square root for several applicable approaches. $\endgroup$
    – dxiv
    Sep 30 '18 at 5:06
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    $\begingroup$ Find $a,b$ such that $\sqrt{9-6\sqrt 2}=\sqrt{a+b-2\sqrt{ab}}$ since $\sqrt{a+b-2\sqrt{ab}}=\sqrt{(\sqrt a-\sqrt b)^2}$. $\endgroup$
    – mathlove
    Sep 30 '18 at 5:07
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Try to use the formula your classmate gave. In this situation, $$9-6\sqrt2={\sqrt3}^2-2{\sqrt{3\times6}}+{\sqrt6}^2\Rightarrow(1)$$ That is because $6{\sqrt2}=2{\sqrt{3\times6}}$ Expression (1) now looks similar to $a^2-2ab+b^2$ where $a=\sqrt3$ and $b=\sqrt6$ Using this we can conclude that $$9-6\sqrt2={\sqrt3}^2-2{\sqrt{3\times6}}+{\sqrt6}^2=(\sqrt3-\sqrt6)^2$$ We can subsitute in the original expression $$\sqrt{9-6\sqrt2}=\sqrt{(\sqrt3-\sqrt6)^2}=-(\sqrt3-\sqrt6)=\sqrt6-\sqrt3$$ The simplest form will be $\sqrt6-\sqrt3$

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Let's forget about $9-6\sqrt{2}$ for a second and just think about the expression your classmate thinks is useful:$$a^2-2ab+b^2.$$

And let's keep in mind our goal here. We're looking for something which is a perfect square (since we want it to play well inside a $\sqrt{\quad}$...

Well, this should remind us of $$a^2\color{red}{+}2ab+b^2=(a+b)^2.$$ But that "$-$" on the $2ab$ term is throwing me off! Is there any way to fix it?

This is where we get something for free from just doing a small change of variable: if we let $c=-b$, we get $$a^2-2ab+b^2=a^2+2ac+c^2.$$ That right hand side is of course just $(a+c)^2$, or better yet $(a-b)^2$. So we now know: $$\color{green}{\sqrt{a^2-2ab+b^2}=a-b}$$ (or rather, fine, $\vert a-b\vert$. FINE.).


That's why what your friend wants to do is reasonable. So nowthe question is: how do we do it?

Ultimately this can just feel like trial-and-error at first, but my instinct here is to say that "$-6\sqrt{2}$" looks a lot like "$-2ab$." Because they both have a minus sign. And $6$ is even. This doesn't work immediately, but when we factor out a $3$ things get much cleaner ...

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\begin{align} 9 - 6\sqrt2 &= 3 (3-2\sqrt2) \\ &= 3((\sqrt2)^2 - 2(1)\sqrt{2} +1^2) \\ &= 3(\sqrt2-1)^2 \end{align} Hence, $$\sqrt{9-6\sqrt2} = \sqrt{3}(\sqrt2 - 1)$$

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Your class mate is being.... clever.

If $\sqrt {9-6\sqrt 2}=a-b $ then $9-6\sqrt 2=a^2-2ab+c^3$

Let $2ab=6\sqrt 2$ and $a^2+b^2=9$.

Can we do that?

If we let $b^2=k $ and $a^2=9-k$ then $ab=\sqrt {k (9-k)}=3\sqrt 2=\sqrt {18} $. Solving $k (9-k)=18$ for $k $ (if it isn't visiblely obvious that we can do it in our heads, it is just a quadratic that we can solve by quadratic formula) and, for covenience, choicing the smaller solution (because we want $a>b $), we get $k =3$ is a good solution..

So $a=\sqrt 6$ and $b=\sqrt 3$.

I.e. in other words

$\sqrt {9-6\sqrt 2}=$

$\sqrt {6-2\sqrt 6\sqrt 3 +3}=$

$\sqrt {(\sqrt 6- \sqrt 3)^2}=$

$|\sqrt 6 - \sqrt 3|=$

$\sqrt 6 -\sqrt 3$.

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The reason for doing that is that $\sqrt{a^2-2ab+b^2} = \sqrt{(a-b)^2} = a-b$. Now try to put your radical in the form your classmate suggested!

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