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Find all real m such that $x^3-2x^2-2x+m$ has 3 distinct rational roots.

Source: School exam paper. No idea why it seems so hard

I don't think rational root theorem works, since m is not necessarily integer (clearly it is rational though)

Assuming $m=\frac{r}{s}$, we need to check $sx^3-2sx^2-2sx+r=0$ for all $s,r$ coprime...

61plus
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    yeah but it doesnt work in this case, as your m is real – 61plus Apr 30 '13 at 17:07
  • Well, if $,m,$ is actually real and not merely an integer then this problem seems to be waaaaaay harder than I thought at first and, undoubtedly, way over the level of any high school I've ever known. – DonAntonio Apr 30 '13 at 17:09
  • does m being rational help – 61plus Apr 30 '13 at 17:12
  • $m$ is clearly rational, and no it doesn't imply is integer. For example $x^2-2x+\frac{3}{4}$ has rational solutions. – N. S. Apr 30 '13 at 17:14
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    If $m$ is rational and in lowest terms $\frac{r}{s}$ where $r, s \in \mathbb{Z}$, multiply through by $s$ to get a polynomial with integer coefficients. – Alex Wertheim Apr 30 '13 at 17:14

2 Answers2

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Let $a, b, c$ be the rational roots. Then we have $$a+b+c=2,\quad ab+bc+ca=-2, \quad m=-abc.$$ Replace $c=2-a-b$, $$a^2+b^2+ab-2(a+b)-2=0.$$ (If $a=b$, then $a$ is irrational) Thus, $$\Delta_b=(a-2)^2-4(a^2-2a-2)=-3a^2+4a+12$$ is a square rational.

The rest is easy.

Added after diner. Since $$-3a^2+4a+12=-3(a+\frac{2}{3})^2+4\times \frac{10}{3}$$ is a square. Let $\frac{3}{2}(a+\frac{2}{3})=\frac{p}{q}$ (suppose $\gcd(p,q)=1$), then $$-3p^2+30 q^2$$ is a square of integer, say $3r$, thus $$-p^2+10q^2=3r^2.$$ It is clear that $2|p-r$. If both are even, then $q$ is even too, contradicting $\gcd(p,q)=1$. So $p$ and $r$ both are odd, then $p^2+3r^2=4 \mod(8)$, therefore $q$ is even, hence $10q^2=0 \mod (8)$, this is an contradiction with $p^2+3r^2=10 q^2$.

If I didn't make mistake, Conclusion: there is no $a, b, c$, there is no $m$.

Ma Ming
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HINT:

Use Repeated Root Theorem to identify the values of $m$ giving repeated roots

lab bhattacharjee
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