1964 AHSME Problems/Problem 30

Problem

The larger root minus the smaller root of the equation $(7+4\sqrt{3})x^2+(2+\sqrt{3})x-2=0$ is

$\textbf{(A) }-2+3\sqrt{3}\qquad\textbf{(B) }2-\sqrt{3}\qquad\textbf{(C) }6+3\sqrt{3}\qquad\textbf{(D) }6-3\sqrt{3}\qquad \textbf{(E) }3\sqrt{3}+2$

Solution

Dividing the quadratic by $7 + 4\sqrt{3}$ to obtain a monic polynomial will give a linear coefficient of $\frac{2 + \sqrt{3}}{7 + 4\sqrt{3}}$ . Rationalizing the denominator gives:

$\frac{(2 + \sqrt{3})(7 - 4\sqrt{3})}{7^2 - 4^2 \cdot 3}$

$\frac{14 - 12 - \sqrt{3}}{1}$

$2 - \sqrt{3}$

Dividing the constant term by $7 + 4\sqrt{3}$ (and using the same radical conjugate as above) gives:

$\frac{-2}{7 + 4\sqrt{3}}$

$-2(7 - 4\sqrt{3})$

$8\sqrt{3} - 14$

So, dividing the original quadratic by the coefficient of $x^2$ gives $x^2 + (2 - \sqrt{3})x + 8\sqrt{3} - 14 = 0$

From the quadratic formula, the positive difference of the roots is $\frac{\sqrt{b^2 - 4ac}}{a}$ . Plugging in gives:

$\sqrt{(2 - \sqrt{3})^2 - 4(8\sqrt{3} - 14)(1)}$

$\sqrt{7 - 4\sqrt{3} - 32\sqrt{3} + 56}$

$\sqrt{63 - 36\sqrt{3}}$

$3\sqrt{7 - 4\sqrt{3}}$

Note that if we take $\frac{1}{3}$ of one of the answer choices and square it, we should get $7 - 4\sqrt{3}$ . The only answers that are (sort of) divisible by $3$ are $6 \pm 3\sqrt{3}$ , so those would make a good first guess. And given that there is a negative sign underneath the radical, $6 - 3\sqrt{3}$ is the most logical place to start.

Since $\frac{1}{3}$ of the answer is $2 - \sqrt{3}$ , and $(2 - \sqrt{3})^2 = 7 - 4\sqrt{3}$ , the answer is indeed $\boxed{\textbf{(D) }}$

1964 AHSC (Problems • Answer Key • Resources)
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1964 AHSME Problems/Problem 30

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