# Question Video: Solving Quadratic Equations with Complex Coefficients Mathematics

Solve 𝑧² + (2 + 𝑖)𝑧 + 𝑖 = 0.

01:28

### Video Transcript

Solve 𝑧 squared plus two plus 𝑖 times 𝑧 plus 𝑖 equals zero.

We use the quadratic formula, substituting the coefficients for 𝑎, 𝑏, and 𝑐. We can simplify under the radical. Two plus 𝑖 squared is two squared, which is four, plus two times- two times 𝑖, which is four 𝑖, plus 𝑖 squared, which is negative one. And from this, we subtract four times one times 𝑖, which is four 𝑖. We see that the terms involving 𝑖 cancel. And we’re left with just four minus one, which is three under the radical.

Notice that we have a positive discriminant here. There’s a bit more simplification to do. Two times one on the denominator is just two. And we can distribute the minus sign over the parentheses. Doing this and rearranging the terms, we get negative two plus or minus root three minus 𝑖 over two. Our solutions are therefore 𝑧 equals negative two plus root three over two minus 𝑖 over two and 𝑧 equals negative two minus root three over two minus 𝑖 over two.

Notice that although our equation had a positive discriminant, we don’t get two real solutions. If a quadratic equation with real coefficients has positive discriminant, then we get two real solutions. However, our quadratic equation has some nonreal coefficients. And so as we’ve seen, we shouldn’t expect it to have two real roots.

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