3 + 2i is the different root Sum of the roots = 3 - 2i + 3 + 2i = 6 manufactured from the roots = (3 - 2i)(3 + 2i) = thirteen Quadratic equation is x^2 - (sum of the roots) x + manufactured from the roots = 0 => x^2 - 6x + thirteen = 0 => b = - 6 and c = thirteen.
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If 3-2i is a root, the other root is 3+2i.
So b is - the sum of the roots = -6
c is the product of the roots = 13.
Answer: x²-6x+13 = 0 is the equation.
3 + 2i is the different root Sum of the roots = 3 - 2i + 3 + 2i = 6 manufactured from the roots = (3 - 2i)(3 + 2i) = thirteen Quadratic equation is x^2 - (sum of the roots) x + manufactured from the roots = 0 => x^2 - 6x + thirteen = 0 => b = - 6 and c = thirteen.
In order for the coefficients to be real, the other root must be 3+2i. See: http://en.wikipedia.org/wiki/Quadratic_equation#Di...
Call the roots r1 and r2. Write the following:
(x - r1) (x - r2) = x² + bx + c
x² - (r1+r2) + (r1×r2)
Thus c = r1 ×r2 and b = - (r1 + r2).
c = r1×r2 = (3 - 2i)(3 + 2i)
= 9 - 6i + 6i -4i² = 9 + 4 = 13
b = - (r1 + r2) = - (3 - 2i + 3 + 2i) = -6
b = -6
c = 13