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

m
(solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
We could multiply the entire [[equation]] by all of the [[denominator]]s, though that would obviously be unnecessarily tedious.
 +
 
 +
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>\displaystyle (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>\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

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

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 $a$ as the denominator of the first fraction. We can rewrite it as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$. Multiplying out the denominators now, we get:

$\displaystyle (a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0$

Simplifying, we get that $-64a + 40 \cdot 16 = 0$, so $a = 10$. Re-substituting the value of $a$, we get that $\displaystyle 10 = x^2 - 10x - 29$. Thus, $\displaystyle 0 = (x - 13)(x + 3)$. The positive root is $013$.

See also

1990 AIME (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS