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The question is asked so that I have multiple choices and need to prove the thing both ways (it's an equivalence).

The problem is, whichever thing I compare to the minimal values of the roots I try to use, whether it be Viete's formula, the Discriminant, or even go the other way and try to solve the equation for $a$, I get one single constraint to $a$, and that is $a>1$, which isn't of any help as all the possible correct answers have some fraction of $9$ or $5$ in them.

What limiting factor am I not taking into consideration? A hint would be nice.

John Doe
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1 Answers1

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The equation can be written as $$(x-3a)^2 + 2-2a = 0 \implies x = 3a \pm \sqrt{2a-2}$$ This forces $a \geq 1$. Hence, the smaller root is $3a - \sqrt{2a-2}$, which needs to exceed $3$, i.e., we need \begin{align} 3a-\sqrt{2a-2} > 3 & \implies 3(a-1) > \sqrt2 \sqrt{a-1} \implies 9(a-1)^2 > 2(a-1)\\ & \implies (a-1) > \dfrac29 \implies a > \dfrac{11}9 \end{align}

Adhvaitha
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  • Thanks for the answer, but I have a question. I have tried at least 3 different approaches to this, including solving for a and then looking at the smaller root (similarly to what you did), why did I never reach the right conclusion? How do I know which way to go around this? – John Doe May 06 '15 at 19:45