Maths Olympiad Prep

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Problem 1078

AMC 12 late, AIME early
Algebra Difficulty 4.7 Prove it Bulgarian Mathematical Competitions · Bulgaria

Find all values of aa such that the equation
(a2−a−9)x2−6x−a=0 (a^{2}-a-9) x^{2}-6 x-a=0
has two distinct positive roots.

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Official solution

Solution:
The equation has two distinct positive roots if and only if
∣ D=a3−a2−9a+9>0x1+x2=6a2−a−9>0x1x2=−aa2−a−9>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∈((1−37)/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,(1−37)/2)a \in(-3,(1-\sqrt{37}) / 2).

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