Question

How do you find the zeros, real and imaginary, of `y=9x^2-7x-3` using the quadratic formula?

Answer 1

`x = 7/18+-sqrt(157)/18`

Explanation:

Given:

`y = 9x^2-7x-3`

Note that this is in standard form:

`y = ax^2+bx+c`

with `a=9`, `b=-7` and `c=-3`

It has zeros which we can find using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-(color(blue)(-7))+-sqrt((color(blue)(-7))^2-4(color(blue)(9))(color(blue)(-3))))/(2(color(blue)(9)))`

`color(white)(x) = (7+-sqrt(49+108))/(18)`

`color(white)(x) = (7+-sqrt(157))/(18)`

`color(white)(x) = 7/18+-sqrt(157)/18`

Note that `157` is prime, so `sqrt(157)` is already in simplest form.