Find the value of $\frac{\sqrt2 + \sqrt6}{\sqrt{2 + \sqrt3}}$

Question

Find the value of $\dfrac{\sqrt2 + \sqrt6}{\sqrt{2 + \sqrt3}}$.

My book is showing that this problem has an exact integer value solutions, but I was unable to find it. How can we prove that it has integer solutions, and which integer is its solution?

Answer 1

I presume you mean $$\frac{\sqrt2+\sqrt6}{\sqrt{2+\sqrt3}}?$$ If you call that $a$, then $$a^2=\frac{2+2\sqrt{12}+6}{2+\sqrt3}=\frac{8+4\sqrt3}{2+\sqrt3}=4$$ so that $a=2$.

Answer 2

Hint: $(1+\sqrt3)^2=2(2+\sqrt3)$