What is $\sqrt {5 - \sqrt {13 + \sqrt {48}}}$?
I can't figure out how to solve this, I don't really know how or where to find the solution, even the same problems just to figure out, but I didn't find anything. Please help. $$ \sqrt {5 - \sqrt {13 + \sqrt {48}}} $$
I was thinking just to use a calculator, but I am not sure about this. Any advices, any links would help very much, Thanks!!!
$$ \left(\sqrt {5 - \sqrt {13 + \sqrt{48}}}\right) = {?} $$ How can I solve this?
$\endgroup$ 63 Answers
$\begingroup$Hint to get you started: $$\sqrt{13 + \sqrt{48}} = \sqrt{(\sqrt{12})^2 + (\sqrt{1})^2 + 2\sqrt{12}\sqrt{1}}$$
$\endgroup$ 2 $\begingroup$The following method is often useful in similar exercises:
Find $a,b$ such that $\left(a+b\sqrt3\right)^2=13+4\sqrt{3}.$ This leads to the system $$\left \{ \begin{array}{1} a^2+3b^2&=13\\ ab&=2 \end{array}\right.$$ with two solutions $(a,b)=(\pm1,\pm2).$ Consequently $\sqrt{13+\sqrt{48}}=1+2\sqrt3.$
Once again this method: $$(c+d\sqrt3)^2=5-\left(1+2\sqrt3\right)$$ leads to $$\left \{ \begin{array}{1} c^2+3d^2&=4\\ cd&=-1 \end{array}\right.$$
and we have $(c,d)=(\pm1,\mp1)$.
We can write $$\sqrt{5-\sqrt{13+\sqrt{48}}}=-1+\sqrt3,$$ as the square root is non-negative.
Since we have $48=3\times 16$, we can write $$\sqrt{5-\sqrt{13+4\sqrt{3}}}$$
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