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Show that if $\alpha$, $\beta$ are the roots of the equation $ax^2 + 3x + 2 = 0,\; a<0$, then $$\dfrac{(\alpha^2)}{(\beta)}+\dfrac{(\beta^2)}{(\alpha)}> 0$$

I could only figure out two things

first, that $a = \frac{-(3\alpha + 2)}{(\alpha^2)}$

and $\frac{(\alpha^2)}{(\beta)}$ + $\frac{(\beta^2)}{(\alpha)}$ = $\frac{(\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta)}{(\alpha\beta)}$.

please help further

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  • $\begingroup$ What are you trying to achieve here? Please update your post and include a question. $\endgroup$ – maxmilgram Mar 13 '19 at 6:29
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Hint

The usual method would be to use the formula $$\alpha \text{ or }\beta=\frac{-3\pm\sqrt{9-8a}}{2a}$$ and then work with this which might be rather boring.

Alternatively, you could observe that $$\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha·\beta}$$

You can compute the numerator with Newton's Sums method...

Using Vieta's Formulas might also be a nice approach ;)

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I don't think the claim is true.

From your last equation, we continue:

Since $\alpha+\beta=-3/a>0$ for $a<0,$ we only need bother to show that $$\frac{\alpha^2+\beta^2-\alpha\beta}{\alpha\beta}$$ is positive. Thus, we get $$\frac{\alpha^2+\beta^2-2\alpha\beta+\alpha\beta}{\alpha\beta}=\frac{(\alpha-\beta)^2+\alpha\beta}{\alpha\beta}=\frac{(\alpha-\beta)^2}{\alpha\beta} + 1.$$ But we have that $$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta+2\alpha\beta-2\alpha\beta=(\alpha+\beta)^2-4\alpha\beta.$$ Therefore we get that $$\frac{\alpha^2+\beta^2-\alpha\beta}{\alpha\beta}=\frac{(\alpha+\beta)^2-3\alpha\beta}{\alpha\beta}.$$ Substituting $-3/a$ for the sum, and $2/a,$ the product in RHS of last equation gives $$\frac{(-3/a)^2-3(2/a)}{2/a}.$$ Clearly the numerator is positive, but the denominator isn't.

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We can write $\alpha\beta=\frac2a$ and $\alpha+\beta=-\frac3a$ by Viète's formulas. Then, with a little help from the binomial theorem, $$\frac{\alpha^2}\beta+\frac{\beta^2}\alpha=\frac{-3/a+((-3/a)^2-3(2/a))}{2/a}$$ $$=\frac{9/a-9}2=\frac92\left(\frac1a-1\right)$$ Since $a<0$, so is $\frac1a-1$, implying that the original expression in $\alpha$ and $\beta$ is also negative.

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