How to find the square root of $11-6\sqrt{2}$?
Can anyone give me some hints for solving this question? I've forgotten how to solve it.
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This is a special case
$$11-6\sqrt{2}=2+9-6\sqrt{2}=(\sqrt{2})^2-2\cdot 3\cdot\sqrt{2}+3^2=(\sqrt{2}-3)^2$$
So square root of $11-6\sqrt{2}$ is $\pm(\sqrt{2}-3)$.
KittyL
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How did you do that? – The Chaz 2.0 Jun 03 '15 at 13:58
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@TheChaz2.0: Consider the formula $(a\pm b)^2=a^2\pm 2ab+b^2$. If this is a special case, the $6\sqrt{2}$ can be written as $2ab=2\cdot 3\sqrt{2}$. Then if $a^2=2, b^2=9$ can add up to $11$, we are done. Which is the case. – KittyL Jun 03 '15 at 14:02
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Right. I understand binomial expansion. What about the square root of $10-6\sqrt(2)$? – The Chaz 2.0 Jun 03 '15 at 14:03
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@TheChaz2.0: This is not a special case, you will have to write $\sqrt{10-6\sqrt{2}}$ then. – KittyL Jun 03 '15 at 14:04
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But I found that $-\sqrt{2}+3$ is also the answer. Is it true? – Mathxx Jun 03 '15 at 14:45
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@Mathxx: Yes, I said "plus or minus". I will change it to $\pm$, if that's better. – KittyL Jun 03 '15 at 14:51
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If I well understand you want: $$ \sqrt{11-6\sqrt{2}} $$ You can use the formula: $$ \sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}} $$ that is usefull if $a^2-b$ is a perfect square.
In your case $a=11$, $b=72$ and it works. You can do it?
Emilio Novati
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I don't know there's a direct formula for this type of question. Great ! – Mathxx Jun 03 '15 at 14:09
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To denest square roots we can use a simple, memorable algorithm described here.
Then $\:11-6\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -7\ $ yields $\ 18-6\sqrt{2}\:$
with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{2\cdot 18}\ =\ 6.\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above we obtain $\ \ \ 3-\sqrt 2$
Bill Dubuque
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Hope that I'll understand this next time. It's too complicated for me. Thanks! – Mathxx Jun 03 '15 at 14:30
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1@Mathxx Please see the linked post for the definition of norm and trace. Once you know that it is simple algebra. – Bill Dubuque Jun 03 '15 at 14:34
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Silly question - why do you say subtracting out the square root of the norm, but then (just) add it? If the norm is $49$, why not just say "add $7 = \sqrt{49}$"? – The Chaz 2.0 Jun 03 '15 at 14:40
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2@TheChaz2.0 Stating the rule that way highlights symmetry. We $\rm\color{#0a0}{subtract}$ out a $\rm\color{#c00}{multiplicative}$ quantity (norm), then $\rm\color{#c00}{divide}$ out an $\rm\color{#0a0}{additive}$ quantity (trace). Subtraction is the additive analog of division. Stating it that way shows the analogy between both steps. The above case amounts to addition only because I chose the negative sqrt $ = -7,$ to simplify the arithmetic We are free to choose the signs in any way - see the linked proof. – Bill Dubuque Jun 03 '15 at 14:59
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