SOLUTION: The roots of the quadratic equation z^2 + az + b = 0 are 2 - 3i and 2 + 3i. What is a+b?

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Question 1085963: The roots of the quadratic equation z^2 + az + b = 0 are 2 - 3i and 2 + 3i. What is a+b?
Found 3 solutions by Boreal, josgarithmetic, ikleyn:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
2+/-3i
b has to be -4, because -(-4)/2, assuming z is 1, is 2.
z^2-4z+b
b^2-4ac has to be -36, because the square root of that is +/-6i and half of that is +/-3i
so, (-4)^2-4(1)c=-36
16-4c=-36
-4c=-52
c=13
z^2-4z+13
a+b=1-4=-3 ANSWER
Answer by josgarithmetic(39615)   (Show Source): You can put this solution on YOUR website!
Not the only method, but





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Answer by ikleyn(52766)   (Show Source): You can put this solution on YOUR website!
.
-a = (2-3i) + (2+3i) = 4  ====>  a = -4.

b = (2-3i)*(2+3i) =  = 4 + 9 = 13.

====>  a + b = -4 + 13 = 9.


The Vieta's theorem.

Answer. a + b = 9.



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