Question

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

Answer 1

`x=(1+-sqrt37)/2`

Explanation:

The quadratic formula find the zeroes of a quadratic equation in the form `y=ax^2+bx+c` through

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

In the given quadratic equation, we know that `a=1,b=-1,c=-9`. Plugging these into the quadratic formula gives

`x=(-(-1)+-sqrt((-1)^2-(4xx-9xx1)))/(2xx1)`

`x=(1+-sqrt(1-(-36)))/2`

`x=(1+-sqrt37)/2`

Thus, the two times when the function equal `0` are at `x=(1+sqrt37)/2` and `x=(1-sqrt37)/2`.