What is the equation of a quadratic with roots #(4-3i)# and #(4+3i)#?

1 Answer
Mar 29, 2018

#x^2-8x+25 = 0#

Explanation:

#x=a# is a zero of a polynomial if and only if #(x-a)# is a factor.

So in our example, we can write:

#(x-(4-3i))(x-(4+3i)) = 0#

Multiply out the left hand side to get this into standard form:

#(x-(4-3i))(x-(4+3i))#

#= x^2-((4-color(red)(cancel(color(black)(3i))))+(4+color(red)(cancel(color(black)(3i)))))x+(4-3i)(4+3i)#

#= x^2-8x+(4^2-(3i)^2)#

#= x^2-8x+(16+9)#

#= x^2-8x+25#

So a suitable quadratic equation is:

#x^2-8x+25 = 0#

A slightly different approach is to note that in general:

#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#

In particular if #alpha# and #beta# are conjugates of the form #p+-qi# then:

#{ (alpha+beta = (p-color(red)(cancel(color(black)(qi))))+(p+color(red)(cancel(color(black)(qi)))) = 2p), (alphabeta = (p-qi)(p+qi) = p^2-(qi)^2 = p^2+q^2) :}#

So we could go straight from zeros #4+-3i# to:

#x^2-2(4)x+(4^2+3^2) = x^2-8x+25#