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How to solve this non-linear system? Get the roots directly.

$ax^3+bx^2+cx+d$

$\frac{b}{a}=-(\alpha+\beta+\gamma)$

$\frac{c}{a}=\alpha\beta+\beta\gamma+\gamma\alpha)$

$\frac{d}{a}=-(\alpha\beta\gamma)$

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For searching purposes: these are the Vieta formulae for polynomial roots. –  J. M. Apr 14 '12 at 11:01

1 Answer 1

Let the roots of $x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}$ be $\{r_1, r_2, r_3\}$. Then $$x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} = (x-r_1)(x-r_2)(x-r_3)$$ Expand the RHS, we get: $$ x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} = {x}^{3}+ \left( -{ r_1}-{ r_3}-{ r_2} \right) {x}^{2}+ \left( {r_1}\,{r_3}+{r_1}\,{ r_2}+{ r_2}\,{ r_3} \right) x-{ r_1}\,{ r_2}\,{ r_3} $$ Equate with the equations you are given, we have: $$r_1 = \alpha, r_2 = \beta, r_3= \gamma.$$

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