Question

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

Answer 1

`x=-1/2+- 1/2sqrt(35)color(white)(.)i`

Explanation:

You have two ways of tackling this. You may complete the square or use 'the formula'

I choose to use the formula. It really is worth trying to commit this one to memory. It crops up quite often.

Standard form `y=ax^2+bx+c color(white)("dd") -> color(white)("dd")x=(-b+-sqrt(b^2-4ac))/(2a)`

Given `y=-x^2-x-9`

where `a=-1;b=-1 and c=-9`

`x=(-(-1)+-sqrt((-1)^2-4(-1)(-9)))/(2(-1))`

`x=(+1+-sqrt(-35))/(-2)`

`x=-1/2+-(sqrt(35)xxsqrt(-1))/2`

`x=-1/2+- 1/2sqrt(35)color(white)(.)i`