# Given the equation 3x^2 +2x +k =0, how do you state the sum and product of the roots?

Jan 2, 2017

The sum of the roots is $- \frac{2}{3}$ and their product is $\frac{k}{3}$

#### Explanation:

Given:

$3 {x}^{2} + 2 x + k = 0$

Divide through by $3$ to get the monic quadratic equation:

${x}^{2} + \frac{2}{3} x + \frac{k}{3} = 0$

Note that if this has roots ${r}_{1}$ and ${r}_{2}$ then:

$0 = \left(x - {r}_{1}\right) \left(x - {r}_{2}\right) = {x}^{2} - \left({r}_{1} + {r}_{2}\right) x + {r}_{1} {r}_{2}$

So equating coefficients, we find that the sum of the roots is $- \frac{2}{3}$ and the product of the roots is $\frac{k}{3}$

$\textcolor{w h i t e}{}$
Footnote

Consider what happens for a polynomial of degree $n$.

When we multiply binomials like this:

$\left(x - {r}_{1}\right) \left(x - {r}_{2}\right) \ldots \left(x - {r}_{n}\right)$

$= {x}^{n}$

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

$+ \left({r}_{1} {r}_{2} + {r}_{1} {r}_{3} + \ldots + {r}_{1} {r}_{n} + {r}_{2} {r}_{3} + \ldots + {r}_{n - 1} {r}_{n}\right) {x}^{n - 2}$

$- \left({r}_{1} {r}_{2} {r}_{3} + \ldots + {r}_{n - 2} {r}_{n - 1} {r}_{n}\right) {x}^{n - 3}$

$+ \ldots \pm {r}_{1} {r}_{2.} . . {r}_{n}$

The coefficients of the resulting polynomial are called the elementary symmetric polynomials in ${r}_{1} , {r}_{2} , \ldots , {r}_{n}$

Any symmetric polynomial in ${r}_{1} , {r}_{2} , \ldots , {r}_{n}$ can be expressed in terms of these elementary ones.

For example, when $n = 2$ we can express ${r}_{1}^{2} + {r}_{2}^{2}$ in terms of the elementary symmetric polynomials ${r}_{1} + {r}_{2}$ and ${r}_{1} {r}_{2}$:

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