Solve three 𝑧 squared plus five
𝑖𝑧 minus two equals zero.
We solve this using the quadratic
formula. The formula tells us that the
quadratic equation 𝑎𝑧 squared plus 𝑏𝑧 plus 𝑐 equals zero has solution 𝑧 equals
negative 𝑏 plus or minus root 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. And this formula applies even if
the coefficients 𝑎, 𝑏, and 𝑐 are complex numbers and not just real numbers. So we can apply it to our
What is negative 𝑏? Well, the coefficient of 𝑧 is five
𝑖. So it’s negative five 𝑖. We then have plus or minus the
square root of 𝑏 squared, which of course is five 𝑖 squared. And from this, we subtract four
times 𝑎, which is three, times 𝑐, which is negative two. And finally, we divide through by
two 𝑎, 𝑎 of course being three.
Now we just need to simplify. We can’t see much with negative
five 𝑖 at the moment. But five 𝑖 squared is negative
25. And four times three times negative
two is negative 24. Two times three is of course
six. And now we can simplify further
under the radical. Negative 25 minus negative 24 is
negative 25 plus 24, which is negative one. And so we can see that we have a
negative discriminant here. And we know what the square root of
negative one is. It’s 𝑖. The two solutions then are 𝑧
equals negative four 𝑖 over six and 𝑧 equals negative six 𝑖 over six. We can simplify the solutions to 𝑧
equals negative two-thirds 𝑖 and 𝑧 equals negative 𝑖.
Note that although we have a
negative discriminant here, our two solutions are not complex conjugates. If a quadratic equation with real
coefficients has a negative discriminant, then we get two complex conjugate
solutions. However, there are no guarantees
when we have a nonreal coefficient as we do here.