Product of Roots of Quadratic Equation

Theorem

Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.

Let $\alpha$ and $\beta$ be the roots of $P$.

Then:

$\alpha \beta = \dfrac c a$

$\ds \alpha$

$=$

$\ds \frac {-b + \sqrt {b^2 - 4 a c} } {2 a}$

$\ds \beta$

$=$

$\ds \frac {-b - \sqrt {b^2 - 4 a c} } {2 a}$

Without loss of generality, selecting $\alpha$ and $\beta$ as such

$\ds \leadsto \ \ $

$\ds \alpha \beta$

$=$

$\ds \frac {\paren {-b - \sqrt {b^2 - 4 a c} } \paren {- b + \sqrt {b^2 - 4 a c} } } {4 a^2}$

$\ds $

$=$

$\ds \frac {b^2 - \paren {b^2 - 4 a c} } {4 a^2}$

$\ds $

$=$

$\ds \frac {4 a c} {4 a^2}$

$\ds $

$=$

$\ds \frac c a$

$\blacksquare$