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

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

@@ Line 4: / Line 4: @@
 == 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 ==
 {{AIME box|year=1990|num-b=3|num-a=5}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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