Difference between revisions of "1964 AHSME Problems/Problem 30"
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<math>\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</math> | <math>\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</math> | ||
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| + | ==Solution== | ||
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| + | Dividing the quadratic by <math>7 + 4\sqrt{3}</math> to obtain a monic polynomial will give a linear coefficient of <math>\frac{2 + \sqrt{3}}{7 + 4\sqrt{3}}</math>. Rationalizing the denominator gives: | ||
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| + | <math>\frac{(2 + \sqrt{3})(7 - 4\sqrt{3})}{7^2 - 4^2 \cdot 3}</math> | ||
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| + | <math>\frac{14 - 12 - \sqrt{3}}{1}</math> | ||
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| + | <math>2 - \sqrt{3}</math> | ||
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| + | Dividing the constant term by <math>7 + 4\sqrt{3}</math> (and using the same radical conjugate as above) gives: | ||
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| + | <math>\frac{-2}{7 + 4\sqrt{3}}</math> | ||
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| + | <math>-2(7 - 4\sqrt{3})</math> | ||
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| + | <math>8\sqrt{3} - 14</math> | ||
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| + | So, dividing the original quadratic by the coefficient of <math>x^2</math> gives <math>x^2 + (2 - \sqrt{3})x + 8\sqrt{3} - 14 = 0</math> | ||
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| + | From the quadratic formula, the positive difference of the roots is <math>\frac{\sqrt{b^2 - 4ac}}{a}</math>. Plugging in gives: | ||
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| + | <math>\sqrt{(2 - \sqrt{3})^2 - 4(8\sqrt{3} - 14)(1)}</math> | ||
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| + | <math>\sqrt{7 - 4\sqrt{3} - 32\sqrt{3} + 56}</math> | ||
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| + | <math>\sqrt{63 - 36\sqrt{3}}</math> | ||
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| + | <math>3\sqrt{7 - 4\sqrt{3}}</math> | ||
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| + | Note that if we take <math>\frac{1}{3}</math> of one of the answer choices and square it, we should get <math>7 - 4\sqrt{3}</math>. The only answers that are (sort of) divisible by <math>3</math> are <math>6 \pm 3\sqrt{3}</math>, so those would make a good first guess. And given that there is a negative sign underneath the radical, <math>6 - 3\sqrt{3}</math> is the most logical place to start. | ||
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| + | Since <math>\frac{1}{3}</math> of the answer is <math>2 - \sqrt{3}</math>, and <math>(2 - \sqrt{3})^2 = 7 - 4\sqrt{3}</math>, the answer is indeed <math>\boxed{\textbf{(D) }}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 00:51, 25 July 2019
Problem
The larger root minus the smaller root of the equation ![\[(7+4\sqrt{3})x^2+(2+\sqrt{3})x-2=0\]](http://latex.artofproblemsolving.com/8/1/9/819ebb25c51b832c43a82b3a979e8a6263293d22.png)
is
Solution
Dividing the quadratic by 
to obtain a monic polynomial will give a linear coefficient of 
. Rationalizing the denominator gives:
Dividing the constant term by (and using the same radical conjugate as above) gives:
So, dividing the original quadratic by the coefficient of 
gives 
From the quadratic formula, the positive difference of the roots is 
. Plugging in gives:
Note that if we take 
of one of the answer choices and square it, we should get 
. The only answers that are (sort of) divisible by 
are 
, so those would make a good first guess. And given that there is a negative sign underneath the radical, 
is the most logical place to start.
Since of the answer is , and 
, the answer is indeed 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
