Difference between revisions of "1990 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
− | We could | + | We could clear out the denominators by multiplying, though that would be unnecessarily tedious. |
− | To simplify | + | To simplify the equation, substitute <math>a = x^2 - 10x - 29</math> (the denominator of the first fraction). We can rewrite the equation as <math>\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0</math>. Multiplying out the denominators now, we get: |
− | + | <cmath>(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0</cmath> | |
− | Simplifying, | + | Simplifying, <math>-64a + 40 \times 16 = 0</math>, so <math>a = 10</math>. Re-substituting, <math>10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3)</math>. The positive [[root]] is <math>\boxed{013}</math>. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:18, 4 July 2013
Problem
Find the positive solution to
Solution
We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute (the denominator of the first fraction). We can rewrite the equation as . Multiplying out the denominators now, we get:
Simplifying, , so . Re-substituting, . The positive root is .
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.