If the A.M. between $a$ and $b$ is twice as great as their G.M., show that $a:b=(2+\sqrt3):(2-\sqrt3)$.
My attempt: $$\frac{a+b}{2}=2\sqrt{ab}$$
$$\sqrt\frac{a}{b}+\sqrt\frac{b}{a}=4$$
Let $\sqrt\frac{a}{b}=x$. So,
$$x+\frac{1}{x}=4$$ $$x^2-4x+1=0$$
$$x=\frac{4\pm\sqrt{16-4}}{2}$$ $$x=2\pm\sqrt3$$
So, $a:b=4+3\pm4\sqrt3=7\pm4\sqrt3$
Yes, you are correct. Just note that $$(2-\sqrt{3})(2+\sqrt{3})=1$$
So $$a:b = (2+\sqrt3)^2:1 = (2+\sqrt3):(2-\sqrt3)$$