What is the equation of a quadratic with roots #(4-3i)# and #(4+3i)#?
1 Answer
Explanation:
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+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
#x^2-2(4)x+(4^2+3^2) = x^2-8x+25#