Difference between revisions of "1990 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
| + | Find the positive solution to | ||
| + | <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. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/SpSuqWY01SE?t=935 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=_vslL2YcEaE | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1990|num-b=3|num-a=5}} | |
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:35, 4 November 2022
Problem
Find the positive solution to
Solution
We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute 
(the denominator of the first fraction). We can rewrite the equation as 
. Multiplying out the denominators now, we get:
![\[(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0\]](http://latex.artofproblemsolving.com/8/5/d/85d76b58a1a9f13d4accc30db88c342522655420.png)
Simplifying, 
, so 
. Re-substituting, 
. The positive root is 
.
Video Solution by OmegaLearn
https://youtu.be/SpSuqWY01SE?t=935
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=_vslL2YcEaE
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
