If 3 + 2i is a solution for x^2 + bx + c = 0, where b and c are real numbers, then what are b and c?
If 3 + 2i is a root then so is 3 - 2i
By Vieta
ax^2 + bx + c = 0
Sum of the roots = -b/a and a =1
So (3 + 2i) + (3 -2i) = 6/1 = -b/1 → b = -6
Product of the roots = c/a
So (3+2i)(3-2i) = 9 - 4i^2 = 9 -4(-1) = 9 + 4 = 13 = 13/1 → 13 = c
