# 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. $$\sqrt {5 - \sqrt {13 + \sqrt {48}}}$$

$$\left(\sqrt {5 - \sqrt {13 + \sqrt{48}}}\right) = {?}$$ How can I solve this?

• What are you trying to figure out, exactly? Whether this number can be written in a simpler form? – Sambo Sep 7 '18 at 12:59
• When you write $\sqrt{4}8$, do you mean $\sqrt{4}\times 8 =16$ or $\sqrt{48}$? – Kusma Sep 7 '18 at 13:00
• @Kusma \sqrt {48}} Sorry i fixed – Asylzat Azaev Sep 7 '18 at 13:01
• Considering the user mentions using a calculator, they apparently want to evaluate the expression to get a decimal. – rschwieb Sep 7 '18 at 13:02
• First write $\sqrt{48} = 2\sqrt{12}$, and later write $\sqrt{12} = 2 \sqrt{3}$. – Daniel Fischer Sep 7 '18 at 13:06

Hint to get you started: $$\sqrt{13 + \sqrt{48}} = \sqrt{(\sqrt{12})^2 + (\sqrt{1})^2 + 2\sqrt{12}\sqrt{1}}$$

• $\sqrt{5 - \sqrt{12} - 1}$ = $\sqrt{3 - \sqrt{12} + 1}$ = $\sqrt3 - 1$ – Asylzat Azaev Sep 7 '18 at 13:28
• Thank you very much, thats really easy, you are the best – Asylzat Azaev Sep 7 '18 at 13:29

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 $$\sqrt{5-\sqrt{13+4\sqrt{3}}}$$