Difference between revisions of "1990 AIME Problems/Problem 4"
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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>. | 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!!!== | ||
| + | https://www.youtube.com/watch?v=_vslL2YcEaE | ||
== See also == | == See also == | ||
Revision as of 08:51, 26 June 2022
Contents
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!!!
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. 
