Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$.

Question

How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?

I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.

Answer 1

Squaring both sides, you get$$3+\sqrt{13+4\sqrt3}=4+2\sqrt3,$$which is equivalent to $\sqrt{13+4\sqrt3}=1+2\sqrt3$. Squaring again both sides, you get $13+4\sqrt3=13+4\sqrt3$, and we're done.

Note that you can square at will, since we are dealing only with numbers greater than or equal to $0$.

Answer 2

$$ \sqrt{3+\sqrt{13+4\sqrt3}}= \sqrt{3+\sqrt{1+2\times2\sqrt{3}+(2\sqrt{3})^2}}$$ $$=\sqrt{3+\sqrt{(1+2\sqrt{3})^2}}$$ $$=\sqrt{3+1+2\sqrt{3}}$$ $$=\sqrt{(1+\sqrt{3})^2}$$ $$=1+\sqrt3$$

Answer 3

By one line:$$\sqrt{3+\sqrt{13+4\sqrt{3}}} =\sqrt{3+\sqrt{1+2\cdot2\sqrt3+12}}= \sqrt{3+1+2\sqrt3}=1+\sqrt3$$

Answer 4

We have $$\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3};$$ that is, $$3+\sqrt{13+4\sqrt{3}}=4+2\sqrt{3},$$ I.e., $$\sqrt{13+4\sqrt{3}}=1+2\sqrt{3},$$ which is $$13+4\sqrt{3}=13+4\sqrt{3}.$$ Now just work backwards.

Answer 5

$$\sqrt{3+\sqrt{13+4\sqrt{3}}}= \sqrt{3+\sqrt{{(2\sqrt{3}+1)}^2}} = \sqrt{4+ 2\sqrt{3}} = \sqrt{{(\sqrt{3}+1)}^2} = \sqrt{3}+1.$$