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

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== Solution ==
 
== Solution ==
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We could multiply the entire [[equation]] by all of the [[denominator]]s, though that would obviously be unnecessarily tedious.
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We could clear out the denominators by multiplying, though that would 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:
+
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:
  
−
:<math>(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0</math>
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<cmath>(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0</cmath>
  
−
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>.
+
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 ==

Revision as of 16:56, 11 April 2008

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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