# Question Video: Solving Quadratic Equations with Complex Coefficients Mathematics

Solve π§Β² β (4 + 4π)π§ + 8π = 0.

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### Video Transcript

Solve π§ squared minus four plus four π π§ plus eight π equals zero.

We use the quadratic formula. We substitute the values of the coefficients for π, π, and π. Under the radical sign, negative four plus four π squared becomes 16 plus 32π minus 16, which is just 32π. And from this, we subtract four times one times eight π, which is 32π, meaning that, under the radical, we have zero. On the denominator, we have two. Writing this again without all the crossings out, we see that our discriminant is zero. The square root of zero is zero. And so we add or subtract nothing, meaning that thereβs only one root. π§ equals four plus four π over two, which is two plus two π.

By the fundamental theorem of algebra, this must be a repeated root. And you can check that it is. So we see that this quadratic has a single repeated nonreal root. It turns out that if the discriminant is zero, then weβre guaranteed a repeated root. If the coefficients of the quadratic are real, then this root must be real itself. But if they are complex, then the roots could be complex.

We can see this by looking at the quadratic formula. If we make the discriminant zero, then weβre left with just π§ equals negative π over two π. This is the value of our repeated root. If the coefficients π and π are real, then the repeated root, negative π over two π, must be real as well. But if π or π or both are nonreal, then negative π over two π could be nonreal as well.