www.mathcentre.ac.uk/.../02-Quadratic-EquationsF61Mb.mp4

In this unit, we'll see how we
can use the imaginary number I

to solve any quadratic equation.

Let's have a look at an example.
Suppose we want to solve the

quadratic equation X squared

minus 2X. Plus 10

equals 0. Now we're going to use
the formula for solving a

quadratic equation. This is the
formula over here and first of

all we need to identify the
values of AB&C to substitute in

the formula. Now the value the
value of a is the coefficient of

X squared, which in this case is

one. The value of B is the
coefficient of X, which is minus

2. And the value of C is the
constant term, which is 10.

Can we substitute these values
into this formula? So here we go

X equals. Minus B, which is
minus minus two, which is +2.

Plus or minus the square root of
B squared, B squared is minus 2

squared, which is +4.

Minus four times a, which was
one and see which was 10.

All divided by 2A and 2A is 2
one or two.

Let's tidy this up. We've got 2
plus or minus. Now let's look

under the square root sign.
We've got 4. Subtract 4 * 1 *

10. 4 * 1 * 10 is 44. Subtract
40 is minus 36, so you'll see

we've got a square root of a
negative number. Here. The

square root of minus 36, and
it's all divided by two.

Let me remind you how you deal
with the square root of a

negative number. The square root
of minus 36. We can write as the

square root of 36 times minus

one. Which is 6.

Times I or six I.

So the square root of this
negative number, the square root

of minus 36, simplifies to just

six I. And finally, if we just
want to tidy this up a bit more,

we can notice that there's a
factor of two in the numerator

and the denominator, which can
be cancelled, which will leave

one plus or minus three. I, so
here we have two solutions of

the quadratic equation. One of
the Solutions is is the number 1

+ 3 I and another is the number
1 - 3 I.

Let's have a look at another
example. In this example, we're

going to study the quadratic
equation, two X squared plus X

Plus One is 0.

Again, in
order to

use the

formula. Which is here we need
to identify the values of AB&C.

The value of a, which is the
coefficient of X squared, is 2.

The value of B is one and the
value of C is also one.

And we substitute these values
into the formula.

So we'll get X equals minus B,
which is minus one plus or minus

the square root of be squared.

Which is 1 squared.

Minus four times a which was two
and see which was one all

divided by. To a which is
2 twos of four.

Let's tidy it. What we've got
minus one plus or minus. Now

let's look at the square root.
We've got 1 squared, which is

one subtract 428, so it's one
subtract 8, which is minus

Seven. You'll see again that
we've ended up with a square

root of a negative number.

Now the square root of minus
Seven we handle in the same way

as before. We write it as the
square root of 7 times minus

one. The square root of Seven
relievers. The square root of 7

and the square root of minus
one. We now right as I.

So this solution we have here
now simplifies to minus one plus

or minus the square root of
minus Seven. We write as the

square root of 7.

Times, I and the whole things
divided by 4 and we can leave

our answer like that. But if we
want to we can write it as

separate terms. We can write it
is minus 1 / 4.

Plus or minus.

The square root of 7.

Divided by 4 multiplied
by I. So either of those

forms are equivalent.

We've now seen how we can
write down the solution of

any quadratic equation.

A number such as this one which
has got a part which is purely

real in this case, minus 1/4 and
a part which is imaginary.

That's the part that this number
here. Route 7 over 4 multiplied

by this imaginary number I a
number such as this is called a

complex number, and in the next
unit will define properly what

we mean by a complex number.