Daily Dose of Solved Questions Sep 1

If $a+b:\sqrt{ab}=4:1$ , where $a>b>0$ , then $a:b$ is

$(a)$ $2+\sqrt{3}:2-\sqrt{3}$
$(b)$ $2-\sqrt{3}:2+\sqrt{3}$
$(c)$ $3+\sqrt{2}:3-\sqrt{2}$
$(d)$ $3-\sqrt{2}:3+\sqrt{2}$

CSAT Mantra