Question 245747
{{{f(x)=x^2+2ix-3}}}
You have it right. The quadratic formula is the way to go. With "a" = 1, "b" = 2i and "c" = -3 we get:
{{{x = (-(2i) +- sqrt((2i)^2 - 4(1)(-3)))/2(1)}}}
Now we simplify:
{{{x = (-(2i) +- sqrt(4i^2 - 4(1)(-3)))/2(1)}}}
Since {{{i^2 = -1}}}:
{{{x = (-(2i) +- sqrt(4(-1) - 4(1)(-3)))/2(1)}}}
{{{x = (-2i +- sqrt(-4 +12 ))/2}}}
{{{x = (-2i +- sqrt(8))/2}}}
Since {{{sqrt(8) = sqrt(4*2) = sqrt(4)*sqrt(2) = 2sqrt(2)}}}:
{{{x = (-2i +- 2sqrt(2))/2}}}
Now we reduce the fraction by canceling common factors. So we factor a 2 out of the numerator:
{{{x = (2(-i +- sqrt(2)))/2}}}
The 2's cancel:
{{{x = -i +- sqrt(2)}}}
or
{{{x = -i + sqrt(2)}}} or {{{x = -i - sqrt(2)}}}
And finally, since we usually write complex numbers with the real part first:
{{{x = sqrt(2) + (-i)}}} or {{{x = -sqrt(2) + (-i)}}}