How do you solve the quadratic with complex numbers given #x^2-4x+5=0#?

1 Answer
Aug 28, 2016

Answer:

#x=2+i# and #x=2-i)#

Explanation:

For the equation #ax^2+bx+c=0#, the roots are given by #x=(-b+-sqrt(b^2-4ac))/(2a)#.

It is apparent that if the discriminant #b^2-4ac<0#, we have complex roots.

In the equation #x^2-4x+5=0#, the discriminant is #(-4)^2-4×1×5=16-20=-4<0# and hence roots are complex.

These are #x=(-(-4)+-sqrt(-4))/(2×1)#

= #(4+-2i)/2# i.e.

#x=2+i# and #x=2-i#