Thread: cubic equation proof needed

Say the equation is

ax3+bx2+cx+d=0

suppose that he roots of this equation are α,β, and γ.
How to prove that,
1) α+β+γ = -b/a
2) αβγ=d/a

Originally Posted by callkalpa

Say the equation is

ax3+bx2+cx+d=0

suppose that he roots of this equation are α,β, and γ.
How to prove that,
1) α+β+γ = -b/a
2) αβγ=d/a

Start by reading this: Vieta's Formulas -- from Wolfram MathWorld

Then use Google.

Hello, callkalpa!

Say the equation is: .$\displaystyle ax^3+bx^2+cx+d\:=\:0$

Suppose that the roots of this equation are: .$\displaystyle \alpha,\:\beta,\:\gamma$

Prove that:

. . $\displaystyle 1)\;\;\alpha + \beta + \gamma \:=\:-\dfrac{b}{a}$
. . $\displaystyle 2)\;\;\alpha\beta\gamma \:=\:-\dfrac{d}{a}$

Divide the equation by $\displaystyle a\!:\;\;x^3 + \dfrac{b}{a}x^2 + \dfrac{c}{a}x + \dfrac{d}{a} \:=\:0$ .[1]


Since $\displaystyle \alpha,\,\beta,\,\gamma$ are roots, the equation has the form:
. . $\displaystyle (x-\alpha)(x-\beta)(x-\gamma) \:=\:0$

which simplifies to:
. . $\displaystyle x^2 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \alpha\gamma)x - \alpha\beta\gamma \:=\:0$ .[2]


Equate coefficients of [1] and [2]:

. . $\displaystyle \begin{array}{ccc}-(\alpha + \beta + \gamma) &=& \dfrac{b}{a} \\ \\[-3mm] \alpha\beta + \beta\gamma + \alpha\gamma &=& \dfrac{c}{a} \\ \\[-3mm] -\alpha\beta\gamma &=& \dfrac{d}{a} \end{array}$


Therefore: .$\displaystyle \begin{Bmatrix}\alpha+\beta+\gamma &=& \text{-}\,\dfrac{b}{a} \\ \\[-3mm] \alpha\beta + \beta\gamma + \alpha\gamma &=& \;\dfrac{c}{a} \\ \\[-3mm] \alpha\beta\gamma &=& \text{-}\,\dfrac{d}{a} \end{Bmatrix}$