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

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

$\Rightarrow \left\{\begin {array} {l}
p=c-b^2\\
q=-bc\\
\end{array}\right.$

So:
$27q^2+4p^3 < 0$

$\Rightarrow 27b^2c^2 +4(c-b^2)^3<0$
This can be expanded and factorised to give:
$(4c-b^2)(c+2b^2)^2<0$

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

$\Rightarrow b^2-4c>0$
$\Rightarrow x^2+bx+c=0$ has two real roots, and hence $x^3+px+q=0$ 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:

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

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

$x^3+px+q=0$

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