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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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/SpSuqWY01SE?t=935
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~ pi_is_3.14
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==Video Solution==
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https://www.youtube.com/watch?v=_vslL2YcEaE
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:35, 4 November 2022

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 $\boxed{013}$.

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

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