### Video Transcript

In this video, we will learn how to
solve quadratic equations whose roots are complex numbers. We will begin by learning how to
solve simple equations which have complex solutions and then look at what the
introduction of the complex number set does to our understanding of the quadratic
formula and the discriminant. Finally, weβll learn how to
reconstruct a quadratic equation given a complex root.

During our exploration of the
concept of numbers, we will have come across equations that have no solutions or at
least those that we presume to have no real solutions. However, the introduction of the
set of complex numbers opens up a whole new world when it comes to these kinds of
equations.

Letβs consider the equation two π₯
squared equals negative eight. To solve this equation, we begin by
dividing through by two. That tells us that π₯ squared is
equal to negative four. We will then find the square root
of both sides of the equation.

In the past, we might have said
that this equation, whose solution is found by finding the square root of a negative
number, doesnβt actually have any real solutions. And of course, this statement would
be entirely correct. However, weβre no longer just
dealing with real numbers. We now have the imaginary number
set. And we recall that these revolve
around the single letter π, where π is defined as the solution to the equation π₯
squared equals negative one. But we often just say that π is
equal to the square root of negative one. And this means we can now solve our
equation by finding the square root of negative four.

To find the square root of a
negative number, we split it up. So, for example, the square root of
negative π is the same as the square root of π multiplied by negative one, which
in turn is equal to the square root of π multiplied by the square root of negative
one. And since the square root of
negative one is π, we see that the square root of negative π is the same as π
root π. In this example, π₯ is equal to
plus or minus π root four. And of course, the square root of
four is two. And we see that the equation two π₯
squared equals negative eight has two solutions: two π and negative two π. Letβs consider another equation
that we would have previously been unable to solve.

Solve the equation five π₯ squared
plus one equals negative 319.

We can begin by solving this
equation just as we would any other, by performing a series of inverse
operations. Weβll start by subtracting one from
both sides of the equation. Five π₯ squared plus one minus one
is simply five π₯ squared. And negative 319 minus one is
negative 320.

Next, we divide through by
five. And we see that π₯ squared is equal
to negative 64. Our final step is to find the
square root of both sides of the equation. The square root of π₯ squared is
π₯. And remember, we can take both the
positive and negative roots of negative 64. And we see that π₯ is equal to plus
or minus the square root of negative 64.

At this point, we choose to rewrite
negative 64 as 64 multiplied by negative one. And we then see that the square
root of negative 64 is the same as the square root of 64 multiplied by the square
root of negative one. The square root of negative one
though is π, and the square root of 64 is eight. And we finished solving our
equation. π₯ has two solutions. Itβs eight π and negative eight
π.

Now in fact we can apply the usual
methods for solving equations to help us solve any equation with nonreal roots. In the case of a quadratic
equation, we might struggle to factorise a quadratic expression. But we can apply the other two
methods weβre confident in. These are the quadratic formula and
completing the square. And there are advantages and
disadvantages to both.

The quadratic formula can be a
little nicer to work with when the coefficient of π₯ squared is not equal to
one. And weβll see in a moment that we
can use part of the formula to help us find the nature and number of roots of the
equation. The completing the square method
though can be fairly efficient when the coefficient of π₯ squared is equal to
one.

Now itβs very much personal
preference. Weβll use the quadratic formula
mainly throughout this video. Then we will look at both methods
during our next example. Letβs remind ourselves of the
quadratic formula and the discriminant. For a quadratic equation given as
ππ₯ squared plus ππ₯ plus π equals zero, where π is not equal to zero, the roots
are given by π₯ equals negative π plus or minus the square root of π squared minus
four ππ all over two π. And we use the discriminant to find
the nature of the roots of the equation. Itβs the part of the formula that
sits inside the square root: π squared minus four ππ.

It makes sense that if the
discriminant is greater than zero, the square root of the discriminant will be a
real number. And this means that there will be
two real roots for our equation. If the value of the discriminant is
equal to zero, there will be exactly one solution. This is known as a repeated
root. It occurs when the turning point of
the curve touches the π₯-axis exactly once. And what about if the value is less
than zero? Well, weβve seen that the square
root of a negative number is not a real number. So there are no real roots. This means that the curve doesnβt
actually intersect the π₯-axis. Letβs look at an example of a
quadratic equation which has nonreal roots.

Solve the quadratic equation π₯
squared minus four π₯ plus eight equals zero. Before we solve this equation, we
can, if we want, double-check the nature of the roots of the equation by finding the
value of the discriminant. Remember, the discriminant of an
equation of the form ππ₯ squared plus ππ₯ plus π equals zero is given by the
formula π squared minus four ππ. And itβs sometimes denoted by this
little triangle.

In our equation, π is the
coefficient of π₯ squared. Itβs one. π is the coefficient of π₯. Itβs negative four. And π is the constant. Itβs eight. The discriminant of our equation is
therefore negative four squared minus four multiplied by one multiplied by eight,
which is negative 16.

We know that if the discriminant is
greater than zero, the equation has two real roots. If itβs equal to zero, it has
exactly one real root. And if itβs less than zero, it has
no real roots. Our discriminant is less than
zero. So the equation π₯ squared minus
four π₯ plus eight equals zero has no real roots.

Knowing that weβre going to get two
complex roots, letβs solve this equation by first looking at the quadratic
formula. The solutions to the equation are
found by negative π plus or minus the square root of π squared minus four ππ all
over two π.

We already saw that π squared
minus four ππ in our example is equal to negative 16. So the solutions to our quadratic
equation are given by negative negative four plus or minus the square root of
negative 16 all over two multiplied by one. That simplifies to four plus or
minus the square root of negative 16 all over two.

And at this stage, weβre going to
rewrite the square root of negative 16 as the square root of 16 multiplied by the
square root of negative one. And this is useful because we know
that the square root of 16 is four and we know that the square root of negative one
is π. So π₯ is equal to four plus or
minus four π all over two. And we can simplify. And we see that the solutions to
the quadratic equation are π₯ equals two plus two π and π₯ equals two minus two
π.

Now in fact this isnβt the only
method for solving this equation. We could have completed the
square. And this is very much personal
preference in an example like this. Letβs see what that wouldβve looked
like.

The first thing we do is halve the
coefficient of π₯. Half of negative four is negative
two. So we write π₯ minus two all
squared. Now negative two squared is
four. So we subtract this four and then
add on that value of eight. And of course, all this is equal to
zero.

We can simplify our equation
somewhat, and we get π₯ minus two all squared plus four equals zero. Weβre going to solve this by
subtracting four from both sides. And that gives us π₯ minus two all
squared equals negative four. Weβre then going to find the square
root of both sides of this equation. The square root of π₯ minus two all
squared is π₯ minus two. And remember, we can take both the
positive and negative roots of negative four. So we see that π₯ minus two is
equal to plus or minus the square root of negative four.

Now using the same method as
earlier, we can see that the square root of negative four is actually the same as
two π. And we can complete this solution
by adding two to both sides of the equation. And once again, we see the
solutions to our equation to be two plus two π and two minus two π.

In fact, itβs no accident that the
roots of the equation are complex conjugates of one another. It really makes a lot of sense,
especially given our second method of solving, that this would be true for any
quadratic equation with complex roots.

We can say that the nonreal roots
of a quadratic equation with real coefficients occur in complex conjugate pairs. And there is actually a lovely
little proof of this. Letβs say we have a quadratic
equation of the form ππ₯ squared plus ππ₯ plus π. Weβre going to let πΌ be a solution
to this equation. And we say that πΌ star is the
complex conjugate to πΌ.

Weβre going to substitute this
complex conjugate into our equation. And when we do, we get π
multiplied by πΌ star all squared plus π multiplied by πΌ star plus π. And here we recall the fact that,
for any two complex numbers, the conjugate of their product is equal to the product
of their conjugates. This means that the square of the
conjugate of our solution is equal to the conjugate of the square.

And we see the first part becomes
π multiplied by πΌ squared star. Since π, π, and π are real
numbers β remember, our quadratic equation has real coefficients and we also know
that the conjugate of a real number is just that number β this can be further
rewritten as shown. And finally, we recall the fact
that, for two complex numbers π§ one and π§ two, the conjugate of their sum is equal
to the sum of their conjugates. And we see that π of πΌ star is
equal to π multiplied by πΌ squared plus π multiplied by πΌ plus π star.

Now we already said that πΌ is a
solution to the equation. This means that ππΌ star plus ππΌ
plus π must be equal to zero. And we, of course, know that the
conjugate of zero is simply zero. And weβve seen that since π of πΌ
star is equal to zero, πΌ star must also be a solution to this equation. And in fact, this is called the
complex conjugate root theorem, and it can be extended into solving polynomials. Letβs have a look at a number of
examples of where this theorem can be used to solve problems involving
quadratics.

The complex numbers π plus ππ
and π plus ππ, where π, π, π, and π are real numbers, are the roots of a
quadratic equation with real coefficients.

Given that π is not equal to zero,
what conditions, if any, must π, π, π, and π satisfy?

In this question, we are told that
π plus ππ and π plus ππ are roots to our quadratic equation with real
coefficients. This equation would usually be of
the form ππ₯ squared plus ππ₯ plus π equals zero, though π, π, and π are not
to be confused with the letters π, π, and π in our complex numbers. So weβll rewrite this as ππ₯
squared plus ππ₯ plus π equals zero.

Now we know that the nonreal roots
of a quadratic equation with real coefficients occur in complex conjugate pairs. And remember, to find the
conjugate, we change the sign of the imaginary part. So the conjugate of π plus ππ is
π minus ππ. And π plus ππ and π plus ππ
must be complex conjugates of one another by this theorem. This means that the conjugate of π
plus ππ must be equal to π plus ππ. So we say that π minus ππ equals
π plus ππ.

And for two complex numbers to be
equal, their real parts must be equal. So here we equate π and π. But their imaginary parts must also
be equal. So we equate the imaginary
parts. And we see that negative π equals
π. So the conditions that π, π, π,
and π must satisfy here is that π must be equal to π and negative π must be
equal to π. This time, weβre going to use our
knowledge of the nature of complex roots of quadratic equations to reconstruct an
equation given one of its roots.

Find the quadratic equation with
real coefficients which has five plus π as one of its roots.

Weβre told that five plus π is a
root of the quadratic equation. And remember, we know that the
nonreal roots of a quadratic equation which has real coefficients occur in complex
conjugate pairs. To find the complex conjugate, we
change the sign of the imaginary part. And we can therefore see that the
roots of our equation are five plus π and five minus π. And this means that our quadratic
equation is of the form π₯ minus five plus π multiplied by π₯ minus five minus π
is equal to zero. And this comes from the fact that
when we solve a quadratic equation by factoring, we equate each expression inside
the parentheses to zero.

So in this case, we would have π₯
minus five plus π equals zero and π₯ minus five minus π equals zero. We would solve this first equation
by adding five plus π to both sides. And we see that π₯ equals five plus
π. And we solve the second equation by
adding five minus π to both sides. And we get that second root π₯
equals five minus π.

Weβre going to need to distribute
these parentheses. Letβs use the grid method here
since thereβs a number of bits and pieces that could trip us up. π₯ multiplied by π₯ is π₯
squared. π₯ multiplied by negative five plus
π is negative π₯ five plus π. Similarly, we get negative π₯
multiplied by five minus π. And negative five plus π
multiplied by negative five minus π gives us a positive five minus π multiplied by
five plus π. And weβll distribute these brackets
using the FOIL method.

Multiplying the first term in the
first bracket by the first term in the second bracket gives us 25. We multiply the outer two terms β
thatβs five π β and the inner two terms β thatβs negative five π. And five π minus five π is
zero. So these cancel each other out. And then we multiply the last
terms. Negative π multiplied by π is
negative π squared. And since π squared is equal to
negative one, we see that these brackets distribute to be 25 minus negative one,
which is equal to 26. So our quadratic equation is
currently of the form π₯ squared minus π₯ multiplied by five plus π minus π₯
multiplied by five minus π plus 26.

We collect like terms. And we get π₯ squared minus five
plus π plus five minus π multiplied by π₯ plus 26. π minus π is zero. And weβre left with π₯ squared
minus 10π₯ plus 26 equals zero.

Now actually there is a formula
that we can use that will save us some time. If we have a quadratic equation
with real roots and a compact solution π plus ππ, the equation of that quadratic
is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero. π is the real part of the
solution. Here thatβs five. And π is the imaginary part. In our solution, thatβs one.

We can substitute what we know
about our complex number into the formula. And we get π₯ squared minus two
times five times π₯ plus five squared plus one squared. Two times five is 10, and five
squared plus one squared is 26. And we see once again we have the
same quadratic equation. And we should be able to see now
why this method can be a bit of a time saver.

In this video, weβve learned that
we can use the quadratic formula or completing the square to solve equations with no
real roots by giving our answers as complex numbers. Weβve also seen that these
solutions occur in complex conjugate pairs. And weβve also learned that we can
reconstruct a quadratic equation given one of its complex solutions. If the solution is π plus ππ,
the quadratic equation is π₯ squared minus two ππ₯ plus π squared plus π squared
equals zero.