Question Video: Solving Quadratic Equations with Complex Coefficients

Solve (2 + 3𝑖)𝑧² + 4𝑧 βˆ’ 6𝑖 + 4 = 0.

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Video Transcript

Solve two plus three 𝑖 times 𝑧 squared plus four 𝑧 minus six 𝑖 plus four equals zero.

We use the quadratic formula. We substitute the values of the coefficients for π‘Ž, 𝑏, and 𝑐. And now we simplify. In the denominator, we get four plus six 𝑖. In the numerator, we get negative four plus or minus a big radical. And inside that radical, four squared is 16. And from that, we subtract four times the product of π‘Ž and 𝑐. When we distribute, we see that the terms involving 𝑖 cancel. And we’re left with just eight plus 18, which is 26. 16 minus four times 26 is negative 88.

Notice that we have a negative discriminant here. The square root of negative 88 is 𝑖 times the square root of 88 or two 𝑖 times the square root of 22. It’s tempting to say that this is our final answer here. But we’d like our denominators to be real if possible. We do that by multiplying the numerator and denominator by the complex conjugate of this denominator.

Let’s do this for the first root. We distribute in the numerator and in the denominator too. And in the denominator, we notice that the terms involving 𝑖 cancel, leaving just the modulus of our complex number in the denominator. That’s four squared plus six squared. We evaluate the denominator and group the real and imaginary parts together in the numerator. And we notice that we can cancel a factor of four. We can therefore write our first root in simplest form as shown. And we can use exactly the same procedure to simplify the second root.

Notice that although our quadratic had a negative discriminant, these two roots are not complex conjugates. They don’t even have the same real parts.

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