Difference between revisions of "1990 AIME Problems/Problem 4"
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To simplify some of the word, a [[substitution]] can be used. Define <math>a</math> as the denominator of the first fraction. We can rewrite it as <math>\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0</math>. Multiplying out the denominators now, we get: | To simplify some of the word, a [[substitution]] can be used. Define <math>a</math> as the denominator of the first fraction. We can rewrite it as <math>\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0</math>. Multiplying out the denominators now, we get: | ||
− | :<math> | + | :<math>(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0</math> |
− | Simplifying, we get that <math>-64a + 40 \cdot 16 = 0</math>, so <math>a = 10</math>. Re-substituting the value of <math>a</math>, we get that <math> | + | Simplifying, we get that <math>-64a + 40 \cdot 16 = 0</math>, so <math>a = 10</math>. Re-substituting the value of <math>a</math>, we get that <math>10 = x^2 - 10x - 29</math>. Thus, <math>0 = (x - 13)(x + 3)</math>. The positive [[root]] is <math>013</math>. |
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=3|num-a=5}} | {{AIME box|year=1990|num-b=3|num-a=5}} | ||
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+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 21:59, 24 November 2007
Problem
Find the positive solution to
Solution
We could multiply the entire equation by all of the denominators, though that would obviously be unnecessarily tedious.
To simplify some of the word, a substitution can be used. Define as the denominator of the first fraction. We can rewrite it as . Multiplying out the denominators now, we get:
Simplifying, we get that , so . Re-substituting the value of , we get that . Thus, . 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 |