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Difference between revisions of "1990 AIME Problems/Problem 4"

 
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== Problem ==
 
== Problem ==
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Find the positive solution to
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<center><math>\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0</math></center>
  
 
== Solution ==
 
== Solution ==
 +
We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
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 +
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>
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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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* [[1990 AIME Problems]]
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{{AIME box|year=1990|num-b=3|num-a=5}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 19:18, 4 July 2013

Problem

Find the positive solution to

$\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0$

Solution

We could clear out the denominators by multiplying, though that would be unnecessarily tedious.

To simplify the equation, substitute $a = x^2 - 10x - 29$ (the denominator of the first fraction). We can rewrite the equation as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$. Multiplying out the denominators now, we get:

\[(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0\]

Simplifying, $-64a + 40 \times 16 = 0$, so $a = 10$. Re-substituting, $10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3)$. The positive root is $\boxed{013}$.

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

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