$$\sqrt{34+15\sqrt2}$$ If we want $34+15\sqrt2$ to be a nice square $(a+b)^2=a^2+2ab+b^2$, most likely it is the case that $15\sqrt2$ corresponds to $2ab$. I don't know what to do from here. Is there any general approach that I can use to determine if it can be further simplified?
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4There's no guarantee that $\sqrt{34+15\sqrt2}$ can be simplified further. Do you have a particular reason to believe that it should be? – DMcMor Jan 19 '21 at 17:09
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2Maybe check your previous question https://math.stackexchange.com/q/3991305/89922 for an approach? – peterwhy Jan 19 '21 at 17:17
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See the nested radical formula. In this case, $34^2-15^2\cdot 2$ is not the square of a rational number, so no further simplification is available. – bjorn93 Jan 19 '21 at 17:32
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Can someone explain to me the downvote? – Math Student Jan 19 '21 at 18:14
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The only possible simplification would be: $$\sqrt{34+15\sqrt{2}}=2^{1/4}\sqrt{\left(1+\sqrt{2}\right)\left(19-\sqrt{8}\right)}$$ – Mourad Jan 19 '21 at 18:23
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Should be $\left(a+\sqrt{2} b\right)^2=34+15 \sqrt{2}$
$a^2+2b^2=34;\;2\sqrt{2}ab=15\sqrt 2\to 2ab=15\to b=\frac{15}{2a} $
$a^2+2\left(\frac{15}{2a}\right)^2=34$
$2 a^4-68 a^2+225=0$
which has no nice solutions, thus $\sqrt{34+15 \sqrt{2}}$ cannot be simplified further.
Raffaele
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