### Video Transcript

In this video, weβre going to see
how to solve quadratic equations with complex coefficients. So how do we solve something of the
form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are allowed to
be complex numbers?

Recall that the discriminant of a
quadratic equation ππ₯ squared plus ππ₯ plus π equals zero is π squared minus
four ππ. The discriminant is often denoted
by capital Ξ. For a quadratic equation with real
coefficients, the discriminant allows us to determine the nature of its roots.

There are three possibilities. If the discriminant is positive,
then we have two real roots. If itβs zero, then we have one
repeated real root. And if itβs negative, then we have
a complex conjugate pair of complex roots. But this is only true if the
quadratic equation has real coefficients, that is, if π, π, and π are all real
numbers.

Weβre going to see what we can say
if π, π, and π are complex numbers. But first, letβs see how to solve a
quadratic equation with complex coefficients. We have the familiar ππ§ squared
plus ππ§ plus π equals zero. But whatβs new is that the
coefficients π, π, and π can be complex numbers.

Now if π, π, and π were real,
then we could just apply the quadratic formula. But it might not be obvious that
this quadratic formula still applies when π, π, and π are complex numbers. So letβs rederive the quadratic
formula by completing the square, checking that each step we use is still valid when
π, π, and π are complex numbers and not just real numbers.

We can divide both sides by π to
make the coefficient of π§ squared one. And now we want to complete the
square, writing π§ squared plus π over ππ§ as a single square. We get π§ plus something
squared. And that something is half the
coefficient of π§, so half of π of π, which is π over two π. And if we distribute, we see that
we do indeed get π§ squared plus π over ππ§. But we also get a term of π
squared over four π squared. We need to subtract this term to
get π§ squared plus π over ππ§. We make this substitution then safe
in the knowledge that all the answer that weβve done holds just as true for complex
numbers as it does for real numbers.

Now we can subtract the constant
terms from both sides. And we can combine the two
fractions on the right-hand side, getting π squared minus four ππ over four π
squared. Now comes a tricky part, we need to
take square roots. It turns out that the answer is
what we expect from our experience with real numbers. We can check by squaring the last
line and observing that we get the line above it. And by the fundamental theorem of
algebra, there canβt be any solutions that weβve missed. We have a degree two polynomial and
so we expect two roots. And these two roots are given by
the plus and minus options on the right-hand side. There canβt be any more
options.

All thatβs left to do is to
subtract π over two π from both sides. And then we get the quadratic
formula that we know and love. π§ equals negative π plus or minus
root π squared minus four ππ all over two π. Letβs see some examples of applying
this formula.

Solve three π§ squared plus five
ππ§ minus two equals zero.

We solve this using the quadratic
formula. The formula tells us that the
quadratic equation ππ§ squared plus ππ§ plus π equals zero has solution π§ equals
negative π plus or minus root π squared minus four ππ all over two π. And this formula applies even if
the coefficients π, π, and π are complex numbers and not just real numbers. So we can apply it to our
problem.

What is negative π? Well, the coefficient of π§ is five
π. So itβs negative five π. We then have plus or minus the
square root of π squared, which of course is five π squared. And from this, we subtract four
times π, which is three, times π, which is negative two. And finally, we divide through by
two π, π of course being three.

Now we just need to simplify. We canβt see much with negative
five π at the moment. But five π squared is negative
25. And four times three times negative
two is negative 24. Two times three is of course
six. And now we can simplify further
under the radical. Negative 25 minus negative 24 is
negative 25 plus 24, which is negative one. And so we can see that we have a
negative discriminant here. And we know what the square root of
negative one is. Itβs π. The two solutions then are π§
equals negative four π over six and π§ equals negative six π over six. We can simplify the solutions to π§
equals negative two-thirds π and π§ equals negative π.

Note that although we have a
negative discriminant here, our two solutions are not complex conjugates. If a quadratic equation with real
coefficients has a negative discriminant, then we get two complex conjugate
solutions. However, there are no guarantees
when we have a nonreal coefficient as we do here.

Solve π§ squared plus two plus π
times π§ plus π equals zero.

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

Notice that we have a positive
discriminant here. Thereβs 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 π over two. Our solutions are therefore π§
equals negative two plus root three over two minus π over two and π§ equals
negative two minus root three over two minus π over two.

Notice that although our equation
had a positive discriminant, we donβt 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 weβve seen, we shouldnβt
expect it to have two real roots.

Solve two plus three π times π§
squared plus four π§ minus six π plus four equals zero.

We use the quadratic formula. We substitute the values of the
coefficients for π, π, and π. And now we simplify. In the denominator, we get four
plus six π. 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 and π. When we distribute, we see that the
terms involving π cancel. And weβre 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
π times the square root of 88 or two π times the square root of 22. Itβs tempting to say that this is
our final answer here. But weβd like our denominators to
be real if possible. We do that by multiplying the
numerator and denominator by the complex conjugate of this denominator.

Letβs 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 π cancel, leaving just the modulus of our complex number
in the denominator. Thatβs 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 donβt even have the same real
parts.

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.

In the examples weβve seen in this
video so far, the discriminant has always been real. But the numbers had to be carefully
chosen to make this the case. In general, the discriminants can
be nonreal. Letβs see an example.

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.

Solve π§ squared plus two minus two
π π§ minus seven plus 26π equals zero.

We use the quadratic formula. We substitute the values of the
coefficients for π, π, and π. We simplify under the radical sign,
noticing thereβs some cancellation. And negative eight π plus 28 plus
104π is 28 plus 96π. We notice that, under the radical,
we have two multiples of four. So we can take out that multiple of
four in each case and bring out a two in front of the radical sign. This allows us to cancel the two in
the denominator. So we get negative one plus π plus
or minus the square root of seven plus 24π.

We use de Moivreβs theorem to find
the square roots. But first, we need to write seven
plus 24π in polar form. Its modulus is 25 and its argument
is arctan 24 over seven. And de Moivreβs theorem for roots
tells us how to find the square roots.

The modulus of the square root is
the square root of the modulus. So thatβs the square root of 25,
which is five. We have to halve the argument. The argument of the square root is
arctan 24 over seven divided by two. Putting these into a calculator, we
find that cos of this value is 0.8 and sin of this value is 0.6. And so upon multiplying by the
modulus five, we add or subtract four plus three π. Therefore, our two roots are π§
equals three plus four π and π§ equals negative five minus two π.

Here, as the discriminant was
complex, we had to use de Moivreβs theorem to find its square roots. Actually, we only found one of its
square roots using de Moivreβs theorem. But we knew that the other must be
its opposite. And we have this plus or minus sign
here.

Here are the key points that weβve
covered in this video. We can solve quadratic equations
with complex coefficients using the quadratic formula. If the discriminant is zero, the
equation has a single repeated root. Unlike for quadratic equations with
real coefficients, this repeated root needs not to be real. If the discriminant is nonzero,
then there are two distinct roots. So we see that the discriminant
tells us whether the quadratic has a repeated root or two distinct roots.

We saw that, for quadratic
equations with real coefficients, the sign of the discriminant told us whether the
roots were real or nonreal. If it was positive, we had two
distinct real roots. If it was zero, we had a single
repeated real root. And if it was negative, we had two
complex conjugate roots. This is not the case for quadratic
equations with complex coefficients. We canβt use the sign of the
discriminant to determine whether the roots are real or nonreal. All we can do is tell whether
thereβs a repeated root or whether there are two distinct roots. And in fact, in general, the
discriminant need not be a real number.