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

1 Answer
Jun 7, 2016

Answer:

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...#