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

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to solve any quadratic equation.

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

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quadratic equation X squared

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minus 2X. Plus 10

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

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

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

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

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X squared, which in this case is

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

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

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

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

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

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squared, which is +4.

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

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

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

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

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

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

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

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

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

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square root of 36 times minus

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one. Which is 6.

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Times I or six I.

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

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of minus 36, simplifies to just

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

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

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

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

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

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

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

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

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Plus One is 0.

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Again, in
order to

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use the

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

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

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

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And we substitute these values
into the formula.

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

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the square root of be squared.

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Which is 1 squared.

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

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

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

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

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

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

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root of a negative number.

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

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

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

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

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

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

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square root of 7.

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

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

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

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Plus or minus.

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The square root of 7.

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Divided by 4 multiplied
by I. So either of those

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forms are equivalent.

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

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any quadratic equation.

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

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

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

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

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

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we mean by a complex number.