Question

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

Answer 1

The sum of the roots is `-2/3` and their product is `k/3`

Explanation:

Given:

`3x^2+2x+k = 0`

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

`x^2+2/3x+k/3 = 0`

Note that if this has roots `r_1` and `r_2` then:

`0 = (x-r_1)(x-r_2) = x^2-(r_1+r_2)x+r_1 r_2`

So equating coefficients, we find that the sum of the roots is `-2/3` and the product of the roots is `k/3`

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Footnote

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

When we multiply binomials like this:

`(x-r_1)(x-r_2)...(x-r_n)`

`=x^n`

`-(r_1+r_2+...+r_n)x^(n-1)`

`+(r_1r_2+r_1r_3+...+r_1r_n+r_2r_3+...+r_(n-1)r_n)x^(n-2)`

`-(r_1r_2r_3+...+r_(n-2)r_(n-1)r_n)x^(n-3)`

`+...+-r_1r_2...r_n`

The coefficients of the resulting polynomial are called the elementary symmetric polynomials in `r_1, r_2,..., r_n`

Any symmetric polynomial in `r_1, r_2, ..., 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_1r_2`:

`r_1^2+r_2^2 = (r_1+r_2)^2 - 2(r_1r_2)`