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Let $\alpha, \beta$ be the roots of $ax^2+bx+c=0$ with $a>0$ where $a, b, c\in \mathbb{R}$. We need to find an equation which roots are $\alpha, -\beta$.

My Trial:

Now $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. Since $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\frac{b^2}{a^2}-4\frac{c}{a}$ therefore $\alpha-\beta=\pm \sqrt{\frac{b^2-4ac}{a^2}}$ and $\alpha(-\beta)=-\frac{c}{a}$.

hence the required equation is $x^2\pm \sqrt{\frac{b^2-4ac}{a^2}}x-\frac{c}{a}=0$ i.e. $ax^2\pm \sqrt{b^2-4ac}x-c=0$.

I got stuck here. I do not know how to identify which sign is appropriate here. Please help.

KON3
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    Hint: $\alpha+\beta=-\frac{b}{a}$ and $\alpha\beta=\frac{c}{a}$. – lisyarus Nov 12 '19 at 10:50
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    There are two solutions in general, because we don't know which of the two roots of the original equation is $\alpha$ and which is $\beta$. – TonyK Nov 12 '19 at 10:51
  • You can't, if $(\alpha,\beta)$ are the roots, so do $(\beta,\alpha)$. In your final equation, one sign is for the pair $(\alpha,-\beta)$, the other one is for $(\beta,-\alpha)$. You need some way to distinguish $\alpha$ from $\beta$ before you can fix the sign. – achille hui Nov 12 '19 at 10:54
  • $f(x)=(x-\alpha)(x-(-\beta))$ should satisfy your requirements. – NoChance Nov 12 '19 at 11:15
  • @TonyK, could please comment on my suggestion in the comment above this one? Thanks. – NoChance Nov 13 '19 at 00:34
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    @NoChance: Take for instance $x^2-3x+2$. Are you going to put $\alpha=1,\beta=2$ in your equation? Or $\alpha=2,\beta=1$? – TonyK Nov 13 '19 at 09:59
  • @TonyK, thanks for your reply, the objective is to find the function. as long as f(x)=(x-1)(x-2)=(x-2)(x-1), it would not matter. It is true, as you said, that we can't specify the exact value for each root from the given information, nevertheless, the function can be constructed from the roots. – NoChance Nov 13 '19 at 12:35
  • @NoChance: You seem to have missed the point entirely. So let me ask you this: what would your answer be for the case $a=1,b=-3,c=2$? – TonyK Nov 13 '19 at 14:09
  • @TonyK, thank you for your time, you may be correct. My understanding is that the parameters $a,b$ and $c$ are determined by comparing the coefficients of the expansion resulting from $(x-\alpha)(x+\beta)=x^2+(\beta-\alpha)x-\alpha\beta$ leading to $a=1$ and $b=\beta-\alpha$ and $c=-\beta\alpha$ as the only possible values for $a,b$ and $c$. By this the quadratic equation is determined. Apologies for wasting your time, you don't have to spend more effort on it. Thanks again. – NoChance Nov 13 '19 at 18:32
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    @NoChance: But what is the new $b$? Is it $3-2$, or $2-3$? (Now do you get it?) – TonyK Nov 13 '19 at 18:44
  • Ahaa! Thanks much I get it now. Well done indeed. – NoChance Nov 13 '19 at 19:28

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