Hello dannee Originally Posted by
dannee 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?
I'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 $\displaystyle c$.$\displaystyle \equiv x^3 +(b-a)x^2+(c-ab)x-ac$
Comparing coefficients:$\displaystyle \left\{\begin {array} {l}
0=b-a\\
p=c-ab\\
q=-ac\\
\end{array}\right.$
$\displaystyle \Rightarrow \left\{\begin {array} {l}
p=c-b^2\\
q=-bc\\
\end{array}\right.$
So:$\displaystyle 27q^2+4p^3 < 0$
$\displaystyle \Rightarrow 27b^2c^2 +4(c-b^2)^3<0$
This can be expanded and factorised to give:
$\displaystyle (4c-b^2)(c+2b^2)^2<0$
$\displaystyle \Rightarrow 4c-b^2<0$
$\displaystyle \Rightarrow b^2-4c>0$
$\displaystyle \Rightarrow x^2+bx+c=0$ has two real roots, and hence $\displaystyle x^3+px+q=0$ has three real roots.
Grandad