Difference between revisions of "1990 AIME Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | {{ | + | We could multiply the entire [[equation]] by all of the [[denominator]]s, though that would obviously be unnecessarily tedious. |
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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: | ||
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| + | :<math>\displaystyle (a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0</math> | ||
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| + | 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>\displaystyle 10 = x^2 - 10x - 29</math>. Thus, <math>\displaystyle 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}} | ||
Revision as of 20:43, 2 March 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 | ||
