1. Originally Posted by ThePerfectHacker
And what are $\displaystyle p\mbox{ and }q$.
$\displaystyle p = -5$
$\displaystyle q = -4$

Originally Posted by ThePerfectHacker
I relized something interesting,
Given $\displaystyle x^3+ax^2+bx+c=0$ which has exactly 1 root.
It is necessary (thought maybe not sufficient) that $\displaystyle b^3=a^3c$.
Can you prove that?
But $\displaystyle b^3$ is not equal to $\displaystyle a^3c$

I really have no idea how to do this. I've never seen a problem like this before, and i dont know how to simplify it. And on top of it all, it only counts 2 marks.

2. Originally Posted by janvdl
But $\displaystyle b^3$ is not equal to $\displaystyle a^3c$
That is because you do not have it in the right form.

$\displaystyle 2x^3 - 5x^2 -4x+3=k$
Thus,
$\displaystyle 2x^3 - 5x^2 - 4x + (3-k)=0$
Thus,
$\displaystyle x^3 - \frac{5}{2}x^2 - 2x+\frac{3-k}{2}=0$
So we require that,
$\displaystyle \left( -\frac{5}{2} \right)\left( \frac{3-k}{2} \right) = (-2)^3$
This gives us an equation for $\displaystyle k$.

And hence the only possible number. You still have to check whether or not it gives you exactly one zero.

3. Hey guys, just thought i'd let you know it went great with my first math paper.

Thank you to everyone who helped.

Oh and TPH, i still owe you thanks. I'll give it to you as soon as i get to my pc.

4. Originally Posted by janvdl
Oh and TPH, i still owe you thanks. .