### 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 thereβs 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 weβre 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 weβre 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.