Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Bulgaria

Problem:
Find all values of aa such that the equation
(a2a9)x26xa=0 (a^{2}-a-9) x^{2}-6 x-a=0
has two distinct positive roots.

Solution

Solution:
The equation has two distinct positive roots if and only if
D=a3a29a+9>0x1+x2=6a2a9>0x1x2=aa2a9>0 \left\lvert\, \begin{aligned} & D=a^{3}-a^{2}-9 a+9>0 \\ & x_{1}+x_{2}=\frac{6}{a^{2}-a-9}>0 \\ & x_{1} x_{2}=\frac{-a}{a^{2}-a-9}>0 \end{aligned} \right.
The first inequality is satisfied for a(3,1)(3,+)a \in(-3,1) \cup(3,+\infty), the second one for a((137)/2,(1+37)/2)a \in((1-\sqrt{37}) / 2,(1+\sqrt{37}) / 2), and the third - for every a(,0)a \in(-\infty, 0). Therefore the required values of aa are a(3,(137)/2)a \in(-3,(1-\sqrt{37}) / 2).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.