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$\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$

I have no idea how to unnest radicals, can anyone help?

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  • $\begingroup$ What are you trying to do with this, simplify it or convert it to a decimal? $\endgroup$
    – Carser
    Mar 7, 2016 at 1:23
  • $\begingroup$ Simplify; I know the answer which is $\sqrt{2} +1$ but no idea how to get it $\endgroup$
    – suomynonA
    Mar 7, 2016 at 1:24
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    $\begingroup$ In general, $a+b\sqrt{d}$ with $a,b,d$ integers might have a square root of the form $a'+b'\sqrt{d}$ only if $a^2-db^2$ is a square. That isn't a sufficient condition, but it is a good start. For example, you need $a+b\sqrt{d}>0$, and it doesn't help you find the square root. Finding the square root is often a trial by error with a little help from knowledge of "Pell-like equations." $\endgroup$ Mar 7, 2016 at 1:42

4 Answers 4

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Here's a hint to give you the kind of idea involved:

The innermost term $3-2\sqrt2$ can be rewritten as $1^2-2\cdot1\cdot\sqrt 2+(-\sqrt{2})^2=(1-\sqrt 2)^2$ which cancels the square root it lies inside. Now simply and apply the same strategy of recognizing squares to keep canceling square roots until you are left with a simple expression.

Edit: How do we recognize these squares? The core idea lies in the "mixed term". Our goal is to make the term inside the current innermost radical look like either $(a+b)^2=a^2+b^2+2ab$ or $(a-b^2)=a^2+b^2-2ab$ which have middle terms $2ab$ or $-2ab$ respectively. So in our example, we had $-2\sqrt 2$, so you know that we're going to get something of the form $(a-b)^2$ with $$ab=\sqrt 2.$$ However, we need $$a^2+b^2=3.$$

At this point you one of the $a$ or $b$ as $\sqrt 2$ since it shows prominently in the $ab$ term. So $a^2+(\sqrt 2)^2=3$, so $a$ needs to be $1$. This checks out with $ab=\sqrt 2$. This last bit, you have to eyeball it a little bit.

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  • $\begingroup$ I see that, but are there any tricks to recognize these squares? $\endgroup$
    – suomynonA
    Mar 7, 2016 at 1:38
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    $\begingroup$ @Anonymous See my edit. Hope this helps! $\endgroup$
    – Arkady
    Mar 7, 2016 at 1:47
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Observe that $$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\qquad\implies\qquad\color{blue}{\sqrt{3-2\sqrt{2}}=\sqrt{2}-1}$$ Then, $$\color{blue}{\sqrt{10+4\sqrt{3-2\sqrt{2}}}}=\sqrt{10+4(\sqrt{2}-1)}=\sqrt{6+4\sqrt{2}}=\sqrt{4+4\sqrt{2}+2}=\color{blue}{\sqrt{(2+\sqrt{2})^2}}$$ So \begin{align} \sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}&=\sqrt{23-6(2+\sqrt{2})}\\ &=\sqrt{11-6\sqrt{2}}\\ &=\sqrt{9-6\sqrt{2}+2}\\ &=\sqrt{(3-\sqrt{2})^2}\\ &=3-\sqrt{2} \end{align} Finally \begin{align} \sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}&=\sqrt{9-2(3-\sqrt{2})}\\ &=\sqrt{3+2\sqrt{2}}\\ &=\sqrt{2}+1 \end{align}

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    $\begingroup$ +1, nice answer :). There is however a little sign mistake in the penultimate inequality: $\sqrt{9-2(3-\sqrt{2})}=\sqrt{3{\color{red}+}2\sqrt{2}}=\sqrt{2}+1$ $\endgroup$
    – Surb
    Sep 23, 2016 at 10:04
  • $\begingroup$ Wow, surprised I didn't notice that $\endgroup$
    – suomynonA
    Sep 25, 2016 at 5:32
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Suppose that the term $\sqrt{3-2\sqrt2}$ can be written as $\sqrt a + b$ for some $a,b\in\mathbb Z$. We can equate these expressions then square both sides: $$\sqrt{3-2\sqrt2} = \sqrt a + b\\ 3 - 2\sqrt2 = a+b^2 + 2b\sqrt a$$ and from here you can see that the solution is $(a,b) = (2,-1)$. This gives you a methodical approach that you can apply recursively.

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Work from the inside out.

Let $\sqrt{3 - 2\sqrt{2}} = \sqrt{a} - \sqrt{b}$, where $a$ and $b$ are rational numbers. Squaring both sides yields $$3 - 2\sqrt{2} = a + b - 2\sqrt{ab}$$ Then \begin{align*} a + b & = 3 \tag{1}\\ -2\sqrt{ab} & = -2\sqrt{2} \tag{2} \end{align*} Dividing equation by $-2$ yields $$\sqrt{ab} = \sqrt{2}$$ Squaring both sides of the equation yields $$ab = 2$$ Hence, $$b = \frac{2}{a}$$ Substituting for $b$ in equation 1 yields \begin{align*} a + \frac{2}{a} & = 3\\ a^2 + 2 & = 3a\\ a^2 - 3a + 2 & = 0\\ (a - 1)(a - 2) & = 0 \end{align*} Hence, $a = 1$ or $a = 2$. If $a = 1$, then $b = 2$. However, $\sqrt{1} - \sqrt{2} = 1 - \sqrt{2} < 0$, but $\sqrt{3 - 2\sqrt{2}} > 0$. Thus, $a = 2$ and $b = 1$. Hence, $$\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$$ Substituting $\sqrt{2} - 1$ for $\sqrt{3 - 2\sqrt{2}}$ yields $$\sqrt{10 + 4\sqrt{3 - 2\sqrt{2}}} = \sqrt{10 + 4(\sqrt{2} - 1)} = \sqrt{6 + 4\sqrt{2}}$$ Let $\sqrt{10 + 4\sqrt{2}} = \sqrt{c} + \sqrt{d}$, where $c$ and $d$ are rational numbers. Continue.

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