# Intermediate value problem

• Feb 13th 2010, 02:41 AM
dannee
Intermediate 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 AM
Condition for a cubic to have real roots
Hello dannee
Quote:

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.

• Feb 13th 2010, 05:27 AM
dannee
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 AM
CaptainBlack
Quote:

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?

Just post the original question, even with your correction this makes no sense.

CB