## anonymous one year ago @ParthKohli Vieta's Formula

• This Question is Open
1. Astrophysics

.

2. ParthKohli

Hello.

3. ParthKohli

Let's first think about a quadratic-polynomial.$ax^2 + bx + c = a\left(x - \alpha\right)\left( x - \beta \right)$$$\alpha, \beta$$ are the roots of the polynomial. Now if we expand the right-hand-side,$ax^2 + bx + c = ax^2 - a(\alpha + \beta )x + a \alpha \beta$Comparing the coefficients,$\alpha + \beta = -\frac{b}{a}$$\alpha \beta = \frac{c}{a}$

4. anonymous

not $$\alpha\beta = \frac{1}{a}$$ ?

5. ParthKohli

no

6. ParthKohli

Now let's think about a cubic.$ax^3 +bx^2 + cx + d = a(x- \alpha) (x - \beta ) (x - \gamma)$$= ax^3 - a(\alpha + \beta + \gamma) x^2+ a( \alpha \beta + \beta \gamma + \gamma \alpha) x - a(\alpha \beta \gamma )$

7. anonymous

And this would work the same way