SOLUTION: I need to find all the zeros of the function f(x)=x^2+2ix-3. I'm trying to use the quadratic equation, but I think I'm doing it wrong. Can someone show me step-by-step how to fin

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I need to find all the zeros of the function f(x)=x^2+2ix-3. I'm trying to use the quadratic equation, but I think I'm doing it wrong. Can someone show me step-by-step how to fin      Log On


   

Question 245747: I need to find all the zeros of the function f(x)=x^2+2ix-3.
I'm trying to use the quadratic equation, but I think I'm doing it wrong.
Can someone show me step-by-step how to find the zeros of this function?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=x%5E2%2B2ix-3
You have it right. The quadratic formula is the way to go. With "a" = 1, "b" = 2i and "c" = -3 we get:
x+=+%28-%282i%29+%2B-+sqrt%28%282i%29%5E2+-+4%281%29%28-3%29%29%29%2F2%281%29
Now we simplify:
x+=+%28-%282i%29+%2B-+sqrt%284i%5E2+-+4%281%29%28-3%29%29%29%2F2%281%29
Since i%5E2+=+-1:
x+=+%28-%282i%29+%2B-+sqrt%284%28-1%29+-+4%281%29%28-3%29%29%29%2F2%281%29
x+=+%28-2i+%2B-+sqrt%28-4+%2B12+%29%29%2F2
x+=+%28-2i+%2B-+sqrt%288%29%29%2F2
Since sqrt%288%29+=+sqrt%284%2A2%29+=+sqrt%284%29%2Asqrt%282%29+=+2sqrt%282%29:
x+=+%28-2i+%2B-+2sqrt%282%29%29%2F2
Now we reduce the fraction by canceling common factors. So we factor a 2 out of the numerator:
x+=+%282%28-i+%2B-+sqrt%282%29%29%29%2F2
The 2's cancel:
x+=+-i+%2B-+sqrt%282%29
or
x+=+-i+%2B+sqrt%282%29 or x+=+-i+-+sqrt%282%29
And finally, since we usually write complex numbers with the real part first:
x+=+sqrt%282%29+%2B+%28-i%29 or x+=+-sqrt%282%29+%2B+%28-i%29