1968 AHSME Problems/Problem 28

Problem

If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$ , then a possible value for the ratio $a/b$ , to the nearest integer, is:

$\text{(A) } 5\quad \text{(B) } 8\quad \text{(C) } 11\quad \text{(D) } 14\quad \text{(E) none of these}$

Solution

$\fbox{D}$

$\frac{a+b}{2}=2\cdot\sqrt{ab}$

$\frac{a}{b} +1=4\cdot\sqrt{\frac{a}{b}}$

setting $x=\sqrt{\frac{a}{b}}$ we get a quadratic equation is $x^2+1=4x$ with solutions $x=\frac{4\pm \sqrt{16-4}}{2}$

$x^2=\frac{a}{b}=(4+3)+4\sqrt{3}~14$ .

1968 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1968 AHSME Problems/Problem 28

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