Video Transcript
Solve the quadratic equation 𝑥
squared minus four 𝑥 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 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero is given by the
formula 𝑏 squared minus four 𝑎𝑐. And it’s sometimes denoted by this
little triangle.
In our equation, 𝑎 is the
coefficient of 𝑥 squared. It’s one. 𝑏 is the coefficient of 𝑥. It’s negative four. And 𝑐 is the constant. It’s 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 it’s equal to zero, it has
exactly one real root. And if it’s less than zero, it has
no real roots. Our discriminant is less than
zero. So the equation 𝑥 squared minus
four 𝑥 plus eight equals zero has no real roots.
Knowing that we’re going to get two
complex roots, let’s solve this equation by first looking at the quadratic
formula. The solutions to the equation are
found by negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all
over two 𝑎.
We already saw that 𝑏 squared
minus four 𝑎𝑐 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, we’re 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 𝑖. So 𝑥 is equal to four plus or
minus four 𝑖 all over two. And we can simplify. And we see that the solutions to
the quadratic equation are 𝑥 equals two plus two 𝑖 and 𝑥 equals two minus two
𝑖.
Now in fact this isn’t 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. Let’s see what that would’ve looked
like.
The first thing we do is halve the
coefficient of 𝑥. Half of negative four is negative
two. So we write 𝑥 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 𝑥 minus two all squared plus four equals zero. We’re going to solve this by
subtracting four from both sides. And that gives us 𝑥 minus two all
squared equals negative four. We’re then going to find the square
root of both sides of this equation. The square root of 𝑥 minus two all
squared is 𝑥 minus two. And remember, we can take both the
positive and negative roots of negative four. So we see that 𝑥 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 𝑖. 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 𝑖 and two minus two 𝑖.
In fact, it’s 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. Let’s say we have a quadratic
equation of the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. We’re going to let 𝛼 be a solution
to this equation. And we say that 𝛼 star is the
complex conjugate to 𝛼.
We’re going to substitute this
complex conjugate into our equation. And when we do, we get 𝑎
multiplied by 𝛼 star all squared plus 𝑏 multiplied by 𝛼 star plus 𝑐. 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.
