SOLUTION: I need to find the zeros of f(x)=x^2+2ix-3. I tried using the quadratic equation, but the i is confusing me. Help?

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need to find the zeros of f(x)=x^2+2ix-3. I tried using the quadratic equation, but the i is confusing me. Help?      Log On


   

Question 245751: I need to find the zeros of f(x)=x^2+2ix-3.
I tried using the quadratic equation, but the i is confusing me.
Help?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Just treat this quadratic as you normally would, but make sure to follow the rules of complex arithmetic.


x%5E2%2B2ix-3=0 Start with the given equation.


Notice that the quadratic x%5E2%2B2ix-3 is in the form of Ax%5E2%2BBx%2BC where A=1, B=2i, and C=-3


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%282i%29+%2B-+sqrt%28+%282i%29%5E2-4%281%29%28-3%29+%29%29%2F%282%281%29%29 Plug in A=1, B=2i, and C=-3


x+=+%28-2i+%2B-+sqrt%28+-4-4%281%29%28-3%29+%29%29%2F%282%281%29%29 Square 2i to get %282i%29%5E2=%282i%29%282i%29=4i%5E2=4%28-1%29=-4


Note: Since i=sqrt%28-1%29, i%5E2=%28sqrt%28-1%29%29%5E2=-1


x+=+%28-2i+%2B-+sqrt%28+-4%2B12+%29%29%2F%282%281%29%29 Multiply -4, 1 and -3 to get -4%281%29%28-3%29=12


x+=+%28-2i+%2B-+sqrt%28+-4%2B12+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-2i+%2B-+sqrt%28+8+%29%29%2F%282%29 Combine like terms.


x+=+%28-2i+%2B-+2%2Asqrt%28+2+%29%29%2F%282%29 Simplify the square root.


x+=+-i+%2B-+sqrt%28+2+%29 Reduce.


x+=+-i+%2B+sqrt%28+2+%29 or x+=+-i+-+sqrt%28+2+%29 Break up the 'plus/minus'


So the solutions are x+=+-i+%2B+sqrt%28+2+%29 or x+=+-i+-+sqrt%28+2+%29