Let's consider the following equation where $m,n$ are real numbers:
$$ x^3+mx+n=0 $$
I need to prove/disprove without calculus that for any real root of the above equation we have that: $$ m^2-4 x_1 n \ge 0$$
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Suppose that $x_1$ is a real root of the cubic, and consider the quadratic equation $x_1x^2+mx+n=0$. This must have $x_1$ as a real solution, so ... ?
Brian M. Scott
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that's a nice trick. :-) – user 1591719 Jul 04 '12 at 20:05
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@Chris' sister: It’s a cute problem. – Brian M. Scott Jul 04 '12 at 20:08
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do you know other solutions excepting this one? Apparently it's easy, but without this trick one may be in trouble. (i think) – user 1591719 Jul 04 '12 at 20:15
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3@Chris' sister: I can’t at the moment think of another. If you’re wondering how I came up with it, I looked at the expression $m^2-4x_1n$ and almost immediately thought discriminant of quadratic. Then everything just fell into place. – Brian M. Scott Jul 04 '12 at 20:19
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in fact, yes, i wanted to know how did you think of it, but now it's clear. The approach is very unusual. – user 1591719 Jul 04 '12 at 20:21