A cubic equation...

• May 11th 2008, 11:02 PM
jerry
A cubic equation...
Problem 243. A cubic equation $\displaystyle x^3+px^2+qx+r=0$ has three roots $\displaystyle x_1, x_2, x_3$ Find: $\displaystyle (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ as an expression containing p, q, and r. This polynomial in p, q, and r is called the discriminant of the cubic equation.

I've mulled over it for some time and seem to get half way there but never quite manage to neatly solve this.
• May 11th 2008, 11:51 PM
earboth
Quote:

Originally Posted by jerry
Problem 243. A cubic equation $\displaystyle x^3+px^2+qx+r=0$ has three roots $\displaystyle x_1, x_2, x_3$ Find: $\displaystyle (x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ as an expression containing p, q, and r. This polynomial in p, q, and r is called the discriminant of the cubic equation.

I've mulled over it for some time and seem to get half way there but never quite manage to neatly solve this.

I don't know if this will help you a little bit further:

$\displaystyle (x-x_1)(x-x_2)(x-x_3)=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2 x_3)x-x_1x_2x_3$

Thus:

$\displaystyle \begin{array}{l}-p= x_1+x_2+x_3 \\ q= x_1x_2+x_1x_3+x_2 x_3 \\ -r= x_1x_2x_3\end{array}$
• May 12th 2008, 02:05 PM
jerry
Thanks but that's in the book also.
I've been doing problems involving Vieta's theorem.

The approach I'm taking to this problem is to expand the brackets, detect the roots, and make replacements accordingly. E.g. $\displaystyle -6a^2b^2c^2 = -6r^2$.

For the other two roots I can only find confusing patterns that make me think I'm close but I can't quite work out what to do with them.

I'm quite sick of this problem and think I've just hit a brick wall.