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In the equation $x^2+2ax+3a=0$ has two solutions $\alpha$ and $\beta$ where $-1<\alpha,\beta<1$. Find out the range of $a$.

I tried to solve it by taking $\alpha^2+\beta^2$. But it does not seem to be a right process.

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Let $$f(x)=(x-\alpha)(x-\beta)=x^2+2ax+3a=(x+a)^2-a^2+3a.$$ We may assume that $\alpha\le \beta$. Then $\alpha\le -a\le \beta$. Since $f$ is strictly decreasing on $(-\infty,-a]$ and strictly increasing on $[-a,+\infty)$, $$-1<\alpha\le\beta<1\iff f(-1)>0, f(-a)\le 0\text{ and } f(1)>0.$$ Note that $f(-1)=a+1$, $f(-a)=-a^2+3a$ and $f(1)=5a+1$. Solving the inequalities you will get that the range of $a$ is $(-\frac{1}{5}, 0]$.

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This can't be the correct answer; if $a=1$, then the roots are imaginary. – Zev Chonoles Aug 2 '13 at 5:26
@ZevChonoles: Thank you. I made some mistake in calculation, and it is fixed now. – 23rd Aug 2 '13 at 5:29

Hint: $\alpha + \beta = -2a, \ \alpha \beta = 3a$. These determine $\alpha$ as a function of $\beta$.

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$$|\alpha|,|\beta|\le 1\implies(1-\alpha^2)(1-\beta^2)\ge 0$$$$\implies1-(\alpha+\beta)^2+2\alpha\beta+(\alpha\beta)^2\ge0$$$$\implies1-4a^2+6a+9a^2\ge 0$$$$\implies(a+1)(5a+1)\ge0$$ Thus , $a\ge-1/5$ or $a\le-1$. But, $-2<-2a=\alpha+\beta< 2\implies -1< a< 1$ also, $a^2\ge 3a$ gives us, $a\le 0$ or $a\ge 3$, we will have, $a\in [-1/5,1)\cap[-\infty, 0]=[-1/5,0]$.

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I think the question is intended to disallow the possibility of $\alpha$ and $\beta$ being non-real. The statement $$-1<\alpha,\beta<1$$ is not a shorthand for $$-1<|\alpha|,|\beta|<1.$$ – Zev Chonoles Aug 2 '13 at 5:32
I never said that, $\implies$ is one directional in first line. Am I right? – Kunnysan Aug 2 '13 at 5:41
I agree with Zev. However, you can get around it by first showing that the condition of real roots is satisfied by requiring $a^2 \geq 3a$. – Calvin Lin Aug 3 '13 at 5:28
$a^2\ge 3a$ condition is already taken care of in last line. And $-1<x<1\iff 0\le |x|<1$, I have used, what's wrong with it? – Kunnysan Aug 3 '13 at 8:02

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