# Cubic equation

• Apr 14th 2014, 05:51 AM
Bude8
Cubic equation
http://puu.sh/88lDI.png

I'm not even sure where to begin... Help will be appreciated :)
• Apr 14th 2014, 06:01 AM
SlipEternal
Re: Cubic equation
\$(x-a)^2(x-b) = x^3+px+q\$

Multiply out:

\$(x-a)^2(x-b) = (x^2-2ax+a^2)(x-b) = x^3-(2a+b)x^2+(2ab+a^2)x-a^2b\$

Equate coefficients:

\$-(2a+b) = 0\$

\$2ab+a^2=p\$

\$-a^2b = q\$

This is a system of three equations in four variables. So, solve for \$p\$ in terms of \$q\$.
• Apr 14th 2014, 06:08 AM
JeffM
Re: Cubic equation
\$x^3 + px + q = (x - a)^2(x - b) = (x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b =\$

\$x^3 - (2a + b)x^2 + a(a + 2b)x - a^2b.\$ Straight forward.

The coefficients of the powers must be equal in the right hand and left expressions.

\$\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.\$

Proceed
• Apr 14th 2014, 11:11 AM
Bude8
Re: Cubic equation
Quote:

Originally Posted by JeffM
\$x^3 + px + q = (x - a)^2(x - b) = (x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b =\$

\$x^3 - (2a + b)x^2 + a(a + 2b)x - a^2b.\$ Straight forward.

The coefficients of the powers must be equal in the right hand and left expressions.

\$\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.\$

Proceed

I'm being really stupid and going wrong somewhere.

http://puu.sh/88G42.png
but I'm not getting that...
• Apr 14th 2014, 11:30 AM
romsek
Re: Cubic equation
Quote:

Originally Posted by JeffM
<snip>
\$\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.\$

Quote:

Originally Posted by Bude8

JeffM and SlipEternal have given you 3 equations that must be satisfied

\$q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.\$

from the last one \$b=-2a\$

substitute that into the first two and you get

\$q=2a^3, ~p=-3a^2\$

\$q^2=4a^6\$

\$\dfrac{q^2}{4}=a^6\$

\$p^3=-27a^6\$

\$-\dfrac{p^3}{27}=a^6\$

\$\dfrac{q^2}{4}=-\dfrac{p^3}{27}\$

\$\dfrac{q^2}{4}+\dfrac{p^3}{27}=0\$

\$27q^2+4p^3=0\$