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Possible Duplicate:
$f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$

How many ordered triples of rational numbers $(e,r,t)$ are there such that the following

$$g(x)=x^3+ex^2+rx+t$$

has roots $e$ and $r$ and $t$?

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    Please share any insight you may have into how these two questions came to be asked within a short time. More generally, please always specify the source of a question if you didn't come up with it yourself.2012-12-09
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    I saw this question posted on another forum2012-12-09

1 Answers 1

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$x^3 + ex^2 +rx + t = (x-e)(x-r)(x-t)$

This gives us the following relations $$e+r+t = -e \implies r+t+2e = 0$$ $$er+rt+et = r \implies r = \frac{-et}{e+t-1}$$ $$ert = -t \implies er =-1 \quad \mbox{or} \quad t = 0$$

Consider the case $e,r,t\neq 0$:

$e = -\frac{1}{r}$. Putting it in the second equation yields $$t = \frac{r(r+1)}{r^2-1} = \frac{r}{r-1}$$ Substituting values of $e,t$ in first relation yields $$r+\frac{r}{r-1}-\frac{2}{r} = 0$$ Simplyfying, $$r^3-2r+2 =0$$ This equation in $r$ has one irrational and two imaginary roots. So, such a solution is not possible.

The other alternatives involve at least one of $e,r,t$ to be $0$.

e = 0

By $ert = -t, t=0. e=0, t=0 \mbox{ gives }r = 0$ by the first relation. So, $(0,0,0)$ is the only solution.

r = 0

$t = 0$ as in the previous case. So, $e=0$ as in the previous case.

t=0

The two non-trivial relations are $er = r, r+2e = 0$. As $r=0$ has already been considered, assume $r\neq 0$. Then, $e = 1, r= -2$. So, (1,-2,0) is a valid solution.

The only two possible solutions are $(0,0,0) \mbox{ and } (1,-2,0)$

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    New answers to duplicates should ideally be posted as answers to the original question in order to keep things in one place. Note that your answer is very similar to my answer to the original question, except you missed the solution with $r=-1$ when you divided through by $r+1$.2012-12-09
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    @joriki: k. i should have checked. my bad2012-12-09
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    I wasn't implying that you should have checked for duplicates yourself; it's just that I'd already pointed out that it's a duplicate a bit before you posted this answer. I guess you didn't see that; no worries.2012-12-09