Question

How do you find the zeros, real and imaginary, of `y=-x^2+4x-42` using the quadratic formula?

Answer 1

The roots are `x_1=2-sqrt(38)i` and `x_2=2+sqrt(38)i`.

Explanation:

Function `y=ax^2+bx+c` has two roots (zeros) `x_{1,2}=(-b+-sqrt(b^2-4ac))/(2a)` or one double root if `b^2=4ac`.

In this case `a=-1,b=4,c=-42`

So you can blindly plug in those values:

#x_{1,2}=(-(4)+-sqrt((4)^2-4(-1)(-42)))/(2(-1))=(-4+-sqrt(16-168))/-2 =(4+-sqrt(-152))/2=(4+-2sqrt(38)i)/2=2+-sqrt(38)i#

The roots are `x_1=2-sqrt(38)i` and `x_2=2+sqrt(38)i`.

`sqrt(38)=6.164414...`