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AIME 1990 · 第 4 题

AIME 1990 — Problem 4

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Find the positive solution to

1x2−10x−29+1x2−10x−45−2x2−10x−69=0\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0
解析

Solution

We could clear out the denominators by multiplying, though that would be unnecessarily tedious.

To simplify the equation, substitute a=x2−10x−29a = x^2 - 10x - 29 (the denominator of the first fraction). We can rewrite the equation as 1a+1a−16−2a−40=0\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(a - 16)(a - 40) + a(a - 40) - 2(a)(a - 16) = 0 Simplifying, −64a+40×16=0-64a + 40 \times 16 = 0, so a=10a = 10. Re-substituting, 10=x2−10x−29⟺0=(x−13)(x+3)10 = x^2 - 10x - 29 \Longleftrightarrow 0 = (x - 13)(x + 3). The positive root is 013\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

Video Solution(Fast, easy)

https://youtu.be/rP8_-36n1ps ~MK