Question 187572: What is the i for in some of the quadratic equations? like one of the answers to a qustion is x=3+-2i.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
is not defined in the real numbers. In order to deal with a radicand that is less than zero, a non-Real number, i (for imaginary), was defined as  . This number, along with any necessary real coefficients, comprise the Complex numbers. Complex numbers are of the form  where i is the imaginary number and a and b are Real numbers.
Any time you use the quadratic formula and the discriminant (the  under the radical) is less than zero, you need to factor out a -1, the square root of which is i, and then you can deal with the real number root of whatever is left under the radical.
The answer you gave,  are the roots of:
I know this because I multiplied:
)(x - (3 - 2i)) = x^2 - (3 + 2i)x - (3 - 2i)x + (3 + 2i)(3 - 2i) = x^2 - 6x + 9 + 4)
So, using the quadratic formula:
 \pm sqrt{36-52}}{2}=\frac{6 \pm sqrt{-16}}{2} = \frac{6 \pm sqrt{(-1)(4)(4)}}{2} = \frac {6 \pm 4i}{2} = 3 \pm 2i)
John
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