1964 AHSME Problems/Problem 30

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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 $\boxed{\textbf{(D) }}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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