Asked by cj
                If a polynomial equation p(x)=0 has 3+4i as a solution 
3-i
4-3i
4+3i
3-4i
            
        3-i
4-3i
4+3i
3-4i
Answers
                    Answered by
            Damon
            
    if 3 + 4 i is a solution
then its complex conjugate
3 - 4 i
is also a solution
because when you solve a quadratic for example
(- b +/- sqrt(b^2-4ac))/2a
if b^2-4ac is negative
then you have
-b + number*i
and
-b - same number*i
They come in complex conjugate PAIRS
    
then its complex conjugate
3 - 4 i
is also a solution
because when you solve a quadratic for example
(- b +/- sqrt(b^2-4ac))/2a
if b^2-4ac is negative
then you have
-b + number*i
and
-b - same number*i
They come in complex conjugate PAIRS
                    Answered by
            John
            
    Are you looking for another solution?
if so what do you think/
    
if so what do you think/
                    Answered by
            Reiny
            
    As Damon said, complex roots come in conjugate pairs.
Using properties of roots:
sum of our roots = 3+4i + 3-4i = 6
product of roots = (3+4i)(3-4i)
= 9 - 16i^2 = 25
p(x) = x^2 - 6x + 25
    
Using properties of roots:
sum of our roots = 3+4i + 3-4i = 6
product of roots = (3+4i)(3-4i)
= 9 - 16i^2 = 25
p(x) = x^2 - 6x + 25
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