### Video Transcript

Solve one plus two π times π§
squared minus three plus π equals zero. Round your answers to three
significant figures.

As thereβs no term involving π§, it
seems a bit silly to break out the quadratic formula. What we can do instead is to
subtract negative three plus π from both sides and then divide both sides by one
plus two π to find π§ squared equals three minus π over one plus two π.

Now we just need to find the square
root of this complex number. But first, we multiply numerator
and denominator by the complex conjugate of the denominator. We can distribute and then
simplify. And we get a fifth minus
seven-fifths π.

Now how do we find the square root
of this number? Well, we can write it in polar form
and then apply de Moivreβs theorem for roots. Whatβs the modulus of this complex
number? We find that it is root two. As our complex number is in the
fourth quadrant, its argument is the arctan of its imaginary part over its real
part. We find that its argument is arctan
of negative seven. And so we can write our complex
number in polar form.

De Moivreβs theorem for roots gives
us the πth roots of a complex number in polar form. Weβre looking for square roots, and
so π is two. Applying this formula to our
example, where π is root two and π is arctan negative seven, we get the
following. Putting these into a calculator, we
get π§ equals 0.898 minus 0.779π and π§ equals negative 0.898 plus 0.77π to three
significant figures.