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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 1

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:

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

$=\frac{14 - 12 - \sqrt{3}}{49-48}$

$=2 - \sqrt{3}$

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

Solution 2

Submitted by BinouTheGuineaPig | A step-by-step solution

The original equation can be manipulated as follows.

$(4+4\sqrt{3}+3)x^2+(2+\sqrt{3})x-2=0$

$(2+\sqrt{3})^2x^2+(2+\sqrt{3})x-2=0$

Substituting $u = (2+\sqrt{3})x$,

$\quad u^2+u-2=0$

$\quad (u-1)(u+2)=0$

$\quad u=1$ or $u=-2$

First root of $x$:

$\quad (2+\sqrt{3})x_1=1$

$\quad x_1=\frac{1}{2+\sqrt{3}}$

$\quad x_1=2-\sqrt{3}$

Second root of $x$:

$\quad (2+\sqrt{3})x_2=-2$

$\quad x_2=\frac{-2}{2+\sqrt{3}}$

$\quad x_2=-2(2-\sqrt{3})$

$\quad x_2=-4+2\sqrt{3}$

Now, to find which root of $x$ is larger:

Assume that

$\quad 2-\sqrt{3}>-4+2\sqrt{3}$, and so

$\quad 6>3\sqrt{3}$

$\quad 2>\sqrt{3}$

$\quad \sqrt{4}>\sqrt{3}$ which is true. Hence, $x_1>x_2$.

Finally, finding the difference between the larger and smaller roots of $x$:

$\quad x_1-x_2$

$\quad =(2-\sqrt{3})-(-4+2\sqrt{3})$

$\quad =6-3\sqrt{3}$

Therefore, the answer is $\boxed{\textbf{(D)}}$.

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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