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
![$\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0$](http://latex.artofproblemsolving.com/f/a/9/fa9c9ff03c23586114f36989a00fe3362d4a8574.png)
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 |