Question

How do you find the zeros, real and imaginary, of `y=x^2 -18x +81` using the quadratic formula?

Answer 1

Only one solution `x=9`
As `("Re"+"Im")-> (9+0i)`

Explanation:

Given: `y=x^2-18x+81`

Using std form of `y=ax^2+bx+c`

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

`a=1`
`b=(-18)`
`c=81`

Thus

`x=(-(-18)+-sqrt((-18)^2-4(1)(81)))/(2(1))`

`x=(18+-sqrt(324-324) )/2`

`x=18/2 =9`