How do you find the quadratic equation with roots at (2-i) and its complex conjugate?

1 Answer
George C.
Jan 9, 2016

Multiply out #(x-(2-i))(x-(2+i))# and simplify to find equation:

#x^2-4x+5 = 0#

Explanation:

The Complex conjugate of #2-i# is #2+i#. You get this by inverting the sign of the imaginary part.

Multiply out #(x-(2-i))(x-(2+i))# using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-2)# and #b=i# as follows:

#(x-(2-i))(x-(2+i))#

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

#=(x-2)^2-i^2#

#=x^2-4x+4+1#

#=x^2-4x+5#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
9955 views around the world
You can reuse this answer
Creative Commons License