$\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}$

Question

$$\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}$$ I'm trying to get the term $$\sqrt{x\pm2\sqrt y}$$ However, I don't know how to.

Answer 1

Set $$x=\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}$$ and square both sides to obtain $$x^2=(3-\sqrt5)+2\sqrt{(3-\sqrt5)(3+\sqrt5)}+(3+\sqrt5)$$ $$x^2=6+2\sqrt{9-5}=6+4=10$$ $$x=\sqrt{10},$$ since clearly $x>0$.

Answer 2

Since $\sqrt{3+\sqrt5}=\frac1{\sqrt2}\sqrt{6+2\sqrt5}=\frac{1+\sqrt5}{\sqrt2}$ and $\sqrt{3-\sqrt5}=\frac{-1+\sqrt5}{\sqrt2}$,$$\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}.$$

Answer 3

$$\sqrt {{3}-\sqrt 5{}} +\sqrt {{3}+\sqrt 5{}} = \sqrt {{\frac52+\frac12}-2\sqrt {\frac52 \frac12}{}} +\sqrt {{\frac52+\frac12}+2\sqrt {\frac52 \frac12}{}} \\ = |\sqrt\frac52 - \sqrt\frac12| + |\sqrt\frac52 + \sqrt\frac12| = \sqrt{10}$$