$2 \cdot 27$ If the arithmetic mean of $a, b$ is equal to twice its geometric mean, where $a>b>0$, then the integer closest to $\frac{a}{b}$ is
(A) 5 .
(B) 8 .
(C) 11 .
(D) 14 .
(E) None of the above.
(19th American High School Mathematics Examination, 1968)
[Solution] Given $\frac{1}{2}(a+b)=2 \sqrt{a b}$, this can be transformed into $\frac{a}{b}+1=4 \sqrt{\frac{a}{b}}$, or $\left(\sqrt{\frac{a}{b}}\right)^{2}-4 \sqrt{\frac{a}{b}}+1=0$.
Solving this, we get $\sqrt{\frac{a}{b}}=2 \pm \sqrt{3}$.
Given $a>b>0$, we know $\frac{a}{b}>1$, and since $2-\sqrt{3}<1$, we discard it. $\therefore \quad \frac{a}{b}=(2+\sqrt{3})^{2}=7+4 \sqrt{3} \approx 14$.
Therefore, the answer is $(D)$.