hey all,

i got the following problem to solve:

i got this term: 27*g^2+4*p^3<0

and i need to proof that there are three real roots to this function:

x^3+px+q=0

is it enough to solve it by min/max points of the function?

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- Feb 13th 2010, 02:41 AMdanneeIntermediate value problem
hey all,

i got the following problem to solve:

i got this term: 27*g^2+4*p^3<0

and i need to proof that there are three real roots to this function:

x^3+px+q=0

is it enough to solve it by min/max points of the function? - Feb 13th 2010, 05:02 AMGrandadCondition for a cubic to have real roots
Hello danneeI'm sure you mean:

$\displaystyle 27q^2+4p^3<0$Then the proof goes like this:

Every cubic has at least one real root, so let's assume that $\displaystyle x^3+px + q = 0 $ has a real root $\displaystyle x = a$. Then:$\displaystyle x^3+px+q \equiv (x-a)(x^2+bx+c)$, for some constants $\displaystyle b$ and $\displaystyle c$.Comparing coefficients:$\displaystyle \equiv x^3 +(b-a)x^2+(c-ab)x-ac$

$\displaystyle \left\{\begin {array} {l}So:

0=b-a\\

p=c-ab\\

q=-ac\\

\end{array}\right.$

$\displaystyle \Rightarrow \left\{\begin {array} {l}

p=c-b^2\\

q=-bc\\

\end{array}\right.$

$\displaystyle 27q^2+4p^3 < 0$This can be expanded and factorised to give:

$\displaystyle \Rightarrow 27b^2c^2 +4(c-b^2)^3<0$

$\displaystyle (4c-b^2)(c+2b^2)^2<0$$\displaystyle \Rightarrow x^2+bx+c=0$ has two real roots, and hence $\displaystyle x^3+px+q=0$ has three real roots.

$\displaystyle \Rightarrow 4c-b^2<0$

$\displaystyle \Rightarrow b^2-4c>0$

Grandad - Feb 13th 2010, 05:27 AMdannee
i'm really really sorry, i wrote a bad syntax.... the right term should be:

$\displaystyle 27^{(q^2)}+4 p^3<0$

and i need to prove that there are three real roots to this function:

$\displaystyle x^3+px+q=0$

thanks alot and sorry again - Feb 13th 2010, 07:14 AMCaptainBlack