solving polynomial

• Sep 28th 2008, 07:28 PM
juldancer
solving polynomial
Find an integer d such that the equation x^3 + 4x^2 - 9x + d = 0 has two roots that are additive inverses of each other.

Does anyone know how to do this?
• Sep 28th 2008, 07:42 PM
Jhevon
Quote:

Originally Posted by juldancer
Find an integer d such that the equation x^3 + 4x^2 - 9x + d = 0 has two roots that are additive inverses of each other.

Does anyone know how to do this?

yup

first recall what additive inverses are. the additive inverse of a number \$\displaystyle a\$ is \$\displaystyle -a\$.

also recall that a cubic where the coefficient of \$\displaystyle x^3\$ is 1 can be written as \$\displaystyle (x - r_1)(x - r_2)(x - r_3) = 0\$ where \$\displaystyle r_1,r_2, \$ and \$\displaystyle r_3\$ are the roots (not necessarily all real) of the equation.

so let \$\displaystyle a\$ and \$\displaystyle -a\$ be the two special roots. call the other root \$\displaystyle b\$. thus we have

\$\displaystyle (x - a)(x + a)(x - b) = 0\$

\$\displaystyle \Rightarrow x^3 - bx^2 - a^2 x + ab = 0\$

now equate coefficients to get the values of \$\displaystyle a\$ and \$\displaystyle b\$, and hence find \$\displaystyle d\$