I have the answer but I need the method. My professor wants it in : a + b √c. He said a hint would be to use algebra.
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4"use algebra"...what a hint x) – Caleb Stanford Sep 26 '16 at 21:38
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What do you mean by "under $a+b\sqrt{c}$? – Mark Fischler Sep 26 '16 at 21:48
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$\sqrt{33+12\sqrt6}=x+y\sqrt6$ (it must be $6$ altought $6a^2$ also but it is unuseful). It implies $2xy=12$ and $x^2+6y^2=33$ hence $(x,y)=(3,2)$ (for positives). – Piquito Sep 26 '16 at 22:07
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1Why close this ? Aside from the fact that there are many questions of this sort. – Rene Schipperus Sep 26 '16 at 23:17
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@ReneSchipperus Aside from the fact that there is no personal input whatsoever in the question, you mean? Dunno... – Did Sep 28 '16 at 16:55
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1@6005 An even better hint would be: use mathematics... – Did Sep 28 '16 at 16:56
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See also: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there), Denesting a square root: $\sqrt{7 + \sqrt{14}}$ (and the posts linked there). – Martin Sleziak Sep 28 '16 at 17:18
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1@Did I dont give a rats ass about personal input. I just solve math problems. – Rene Schipperus Sep 29 '16 at 00:04
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@ReneSchipperus Good to know, but then why ask a reason why other users follow the rules of the site? – Did Sep 29 '16 at 05:39
4 Answers
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Hint: Try to solve
$$33+12\sqrt{6} = (a+\sqrt{x})^2$$
by setting the parts without a square root equal and the parts with a square root equal.
For example, to solve:
$$\sqrt{7+4\sqrt{3}} = a+\sqrt{x}$$
Begin with
$${7+4\sqrt{3}} = (a+\sqrt{x})^2$$
$$7+4\sqrt{3} = a^2+x+2a\sqrt{x}$$
$$a^2+x=7,a\sqrt{x}=2\sqrt{3}$$
$$\vdots$$
tomi
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Carl Schildkraut
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Not to be taken too seriously, in the sense that guessing at the values can be more fun, but here's a general formula $$ \sqrt{a+\sqrt{b}}= \sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}} $$ that can be proved by squaring. Indeed, the square of the right-hand side is $$ \frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+ 2\sqrt{\frac{a^2-(a^2-b)}{4}}=a+\sqrt{b} $$ (A different matter would be how to find that formula; I learned it in high school, our books had plenty of exercises where to apply it.)
In your case $a=33$ and $b=12^2\cdot 6=864$. Now $$ a^2-b=33^2-864=1089-864=225=15^2 $$ so the formula tells us that $$ \sqrt{33+12\sqrt{6}}= \sqrt{\frac{33+15}{2}}+\sqrt{\frac{33-15}{2}}= \sqrt{24}+\sqrt{9}=2\sqrt{6}+3 $$
There is a companion formula: $$ \sqrt{a-\sqrt{b}}= \sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}} $$
egreg
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It looks like you mean to ask about $$ \sqrt{33+12\sqrt{6}} $$ You can't tell that from the way you stated the problem.
At any rate, you expand $(a + b\sqrt{6})^2$ and get $$ (a^2+6b^2) + 2ab\sqrt{6} $$ And if this is going to look like $33+12\sqrt{6}$ with $a$ and $b$ integers, then $ab=6$ and there are only $4$ (or 8 including negatives) pairs to try. You find the answer $3+2\sqrt{6}$.
Mark Fischler
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By squaring,
$$\sqrt{33+12\sqrt6}=a+b\sqrt c\Leftarrow33+12\sqrt6=a^2+b^2c+2ab\sqrt c.$$
Then you need to solve
$$\begin{cases}a^2+b^2c=33\\ab\sqrt c=6\sqrt6.\end{cases}$$
If we assume $c=6$, we can try the factors of $6$ for $ab$:
$$1\cdot6\to1^2+6^26\ne33$$ $$2\cdot3\to2^2+3^26\ne33$$ $$3\cdot2\to3^2+2^26=33$$
and we are done.