# What is the product of the roots of ax^2+bx+c = 0" " where a != 0 ?

May 15, 2016

$\frac{c}{a}$

#### Explanation:

Method 1 - Direct multiplication

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

$= \frac{{\left(- b\right)}^{2} - {\left(\sqrt{{b}^{2} - 4 a c}\right)}^{2}}{4 {a}^{2}}$

$= \frac{{b}^{2} - \left({b}^{2} - 4 a c\right)}{4 {a}^{2}}$

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

$= \frac{c}{a}$

$\textcolor{w h i t e}{}$
Method 2 - Think about the roots

Call the two values of this expression ${r}_{1}$ and ${r}_{2}$ - the roots of the quadratic equation $a {x}^{2} + b x + c = 0$.

Then we have:

$a {x}^{2} + b x + c$

$= a \left(x - {r}_{1}\right) \left(x - {r}_{2}\right)$

$= a \left({x}^{2} - \left({r}_{1} + {r}_{2}\right) x + {r}_{1} {r}_{2}\right)$

$= a {x}^{2} - a \left({r}_{1} + {r}_{2}\right) x + a {r}_{1} {r}_{2}$

Equating coefficients we find:

$c = a {r}_{1} {r}_{2}$

So dividing both sides by $a$ we find:

${r}_{1} {r}_{2} = \frac{c}{a}$