Question Video: Solving Quadratic Equations with Complex Coefficients Mathematics

Solve 3𝑧² + 5𝑖𝑧 βˆ’ 2 = 0.

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Video Transcript

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 π‘Žπ‘§ squared plus 𝑏𝑧 plus 𝑐 equals zero has solution 𝑧 equals negative 𝑏 plus or minus root 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž. And this formula applies even if the coefficients π‘Ž, 𝑏, and 𝑐 are complex numbers and not just real numbers. So we can apply it to our problem.

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 π‘Ž, which is three, times 𝑐, which is negative two. And finally, we divide through by two π‘Ž, π‘Ž 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.

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