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?

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- Sep 28th 2008, 07:28 PMjuldancersolving 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 PMJhevon
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 $\displaystyle d$