### Video Transcript

In this video, we will learn how to
solve quadratic equations using the quadratic formula. This is a standard formula that we
can apply to help us find the solutions to certain types of quadratic equations. Suppose we have the general
quadratic equation ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π
are constants, that is, just numbers, and π isnβt equal to zero. The reason for this is that if π
was zero, we would just have a linear equation. The quadratic formula tells us that
the solutions to this quadratic equation, if they exist, are given by π₯ equals
negative π plus or minus the square root of π squared minus four ππ all over two
π.

Now, when I said if they exist, I
mean that there are certain quadratic equations which actually have no real-valued
solutions. And trying to use this formula
wonβt work. Can you see from the formula how
this might be the case? Well, we can see that the formula
includes the square root of π squared minus four ππ. And we also know that whenever we
take a square root, we need the value weβre attempting to square-root to be either
positive or zero. So, if the expression underneath
the square root β thatβs π squared minus four ππ β turns out to have a negative
value, weβd be trying to find the square root of a negative value, which wonβt
work. Now, you donβt need to worry about
this now because the quadratic equations youβre asked to solve at this level wonβt
have this problem. But if you go on to study maths at
a higher level, youβll look at this in a little bit more detail.

Now, if youβre interested in where
this formula comes from, it can actually be derived by rearranging the equation ππ₯
squared plus ππ₯ plus π equals zero to make π₯ the subject. And Iβve written the first couple
of steps in this process here. One of the first key steps is that
we need to factor by π. Thatβs the coefficient of π₯
squared. And, as we said, π must be
nonzero. We can then divide both sides of
the equation by this factor. The next key step is that we need
to use the technique of completing the square. And Iβve written the first stage of
this here. If you fancy a challenge, itβs a
great exercise to practice your algebra skills to work through the remainder of this
method.

Now, itβs also important to point
out that if a quadratic equation can be solved by factorizing, then using the
quadratic formula will still work and will give exactly the same result. Although it will probably be more
effort for you and thereβs more chance of making a mistake. Where the quadratic formula is
really useful is when a quadratic equation cannot be factorized. The focus of this video will be
gaining confidence in applying this formula. Now, depending where you are in the
world and the requirements of your local exam board, you probably need to know this
formula off by heart. Donβt worry if it looks a little
bit scary at this stage. The more times you use it, the more
natural it will become. And youβll be surprised how quickly
you learn the formula. Donβt believe me? Well, letβs try some examples.

Find the solution set of the
equation five π₯ squared minus seven π₯ minus 32 equals zero, giving values to three
decimal places.

So, we have the quadratic equation
and weβre asked to find its solution set, which is just another way of saying solve
the equation. Weβre told that we need to give
values to three decimal places. So, this is a big clue that the
quadratic equation weβve been given wonβt be able to be solved by factoring. And so, we need to use another
method. This is where we need to use the
quadratic formula, which we should make sure we learn off by heart. The quadratic formula tells us that
if we have the general quadratic equation ππ₯ squared plus ππ₯ plus π equals
zero. Then the solution set or roots of
this equation, if they exist, are given by π₯ equals negative π plus or minus the
square root of π squared minus four ππ all over two π.

In order to apply this formula
then, what we need to do are determine the values of π, π, and π. π is the coefficient of π₯
squared. So, for our equation, thatβs
five. π is the coefficient of π₯. So, for our equation, thatβs
negative seven. And we must make sure that we
include the negative sign. π is the constant term. In our equation, thatβs negative
32. Now, all we need to do is
substitute these values of π, π, and π into the quadratic formula. So, we have π₯ equals negative π β
thatβs negative negative seven β plus or minus the square root of π squared, so
negative seven squared, minus four ππ β thatβs minus four times five times
negative 32. And this is all over two π. Thatβs two times five. You can break the formula down into
stages if you wish, or you can substitute straight into the formula. But you must be careful.

Letβs now look at simplifying. We have negative negative
seven. So thatβs seven. And in the denominator, two times
five is 10. Letβs now look at the square
root. Firstly, we have negative seven
squared, and negative seven squared is equal to 49. If youβre using a calculator to
help with this, itβs really common to make a mistake when finding this value. Because what a lot of people do by
mistake is they type negative seven squared into their calculator, exactly as Iβve
written in orange here.

Now, when you do this, your
calculator will give negative 49, but that doesnβt mean your calculator has made a
mistake. It means youβve made a mistake in
the question youβve asked it. Negative seven squared means
negative one multiplied by seven squared. And if you recall BIDMAS or the
order of operations, we work our indices or powers before multiplication. So, negative seven squared means
negative one multiplied by seven squared, which is negative one multiplied by 49,
giving negative 49. What we shouldβve typed into our
calculator was negative seven enclosed in brackets squared. So that ensures that our calculator
is taking a negative value first and then squaring it. This is a really common mistake,
and one you need to make sure you watch out for.

Continuing under the square root
then, we are subtracting four times five times negative 32. Now, four times five times negative
32 is negative 640. So we are subtracting negative 640,
which means overall weβre adding 640. But again, you need to be really
careful with the negative signs here. Simplifying under the square root,
49 plus 640 is 689. So, we have that π₯ is equal to
seven plus or minus the square root of 689 over 10.

Now, weβre asked to give our values
as decimals, so we now need to use our calculator to evaluate each of these. The first root or value in our
solution set is seven plus the square root of 689 all over 10, which is equal to
3.32488 continuing. The second root or second value in
our solution set is seven minus the square root of 689 all over 10. And thatβs equal to negative
1.92488. Weβre asked to give the values to
three decimal places, and in each case, the fourth decimal place is an eight. So, we round up. We can then give our two roots in
set notation. The solution set to this equation
is 3.325, negative 1.925.

The important stages in this
question then were correctly identifying the values of π, π, and π from our
quadratic equation, including their signs, and then carefully working with the signs
when applying the quadratic formula. In our next example, weβll see how
to apply the quadratic formula to an equation which may not look like a quadratic
equation at first glance.

Find the solution set of the
equation two π₯ minus five equals six over π₯, giving values to three decimal
places.

Now, at first glance, we may not
know how to solve this equation as it involves a reciprocal term, six over π₯. Thereβs a trick, though, that we
need to spot, which is that if we multiply the entire equation by π₯, this will
eliminate the fraction. On the left-hand side, weβll have
the π₯ multiplied by two π₯ minus five. And on the right-hand side, π₯
multiplied by six over π₯. We can distribute the parentheses
or expand the brackets on the left-hand side to give two π₯ squared minus five
π₯. And on the right-hand side, π₯
multiplied by six over π₯ simplifies to six. And weβve eliminated the
fraction. We now see that what we have is a
quadratic equation. And so, our next step is to collect
all the terms on the same side.

We can do this by subtracting six
from each side of the equation, giving two π₯ squared minus five π₯ minus six is
equal to zero, a quadratic equation in its most easily recognizable form. Now, we could perform a quick check
to see whether this quadratic equation can be solved by factoring. But as weβre told in the question
to give the solution set to three decimal places, thatβs a huge clue that it
canβt. So, weβre going to need to apply
the quadratic formula instead. This tells us that, for the general
quadratic equation ππ₯ squared plus ππ₯ plus π equals zero. Then the solution set or the roots
of this quadratic equation are given by π₯ equals negative π plus or minus the
square root of π squared minus four ππ all over two π.

Now, itβs important to notice that
the quadratic formula uses a quadratic equation in the form where all the terms are
on one side, and we have zero on the other. So, when weβre identifying the
values of π, π, and π to substitute into the formula, we must use our quadratic
equation in its rearranged form, not in its earlier form of two π₯ squared minus
five π₯ equals six. If we were to use the earlier form,
then we would have the sign for the value of π incorrect, which would lead to an
incorrect solution. Letβs determine then the values of
π, π, and π for our quadratic equation. π is the coefficient of π₯
squared. Thatβs equal to two. π is the coefficient of π₯, which
is equal to negative five. And finally, π is the constant
term, which is equal to negative six. Remember, we must include the sign
with each of these values.

Next, we need to substitute the
values of π, π, and π carefully into the quadratic formula. We have π₯ equals negative π β
thatβs negative negative five β plus or minus the square root of π squared,
negative five squared, minus four ππ β thatβs minus four times two times negative
six β all over two π, which is two times two. Now, we simplify, being very
careful with the negative signs. Negative negative five is simply
five. And in the denominator, two times
two is four. Within the square root, negative
five squared is 25.

Be very careful here, particularly
if youβre using a calculator. A common mistake is to write that
as negative 25. Weβre then subtracting four times
two times negative six. So, weβre subtracting negative
48. Within the square root, 25 minus
negative 48 is the same as 25 plus 48, which is 73. So, we have that π₯ is equal to
five plus or minus the square root of 73 over four. And thatβs our exact answer in
terms of surds.

The question though asked us to
give our values to three decimal places. So, we now need to evaluate these
on a calculator. Our first value is π₯ equals five
plus the square root of 73 all over four, which is 3.38600. Our second value is five minus the
square root of 73 all over four, which is negative 0.88600. Rounding these values to three
decimal places then gives the solution set of 3.386, negative 0.886.

There are two important points to
remember from this question. Firstly, we must make sure our
quadratic equation is in the correct form with all the terms on the same side of the
equation before identifying the coefficients π, π, and π. And secondly, we must be really
careful with negative values in our quadratic formula. Letβs consider one final example,
which will reveal something interesting about the relationship between the sum of
the roots of a quadratic equation and its coefficients.

The sum of the roots of the
equation four π₯ squared plus ππ₯ minus four equals zero is negative one. Find the value of π and the
solution set of the equation.

So, weβre looking, first of all, to
find the value of π, which is a missing coefficient in this quadratic equation. Itβs the coefficient of π₯. Weβre going to need to use the
quadratic formula to find expressions for the roots of this equation, which will be
in terms of π. And we recall that, for the general
quadratic equation ππ₯ squared plus ππ₯ plus π equals zero, its roots are given
by π₯ equals negative π plus or minus the square root of π squared minus four ππ
all over two π.

Letβs determine the values of π,
π, and π for this quadratic equation then, which is in the correct form with all
the terms on the same side. The value of π is the coefficient
of π₯ squared. So π is equal to four. The value of π is the coefficient
of π₯. So π is equal to this unknown
value π. And the value of π is the constant
term, so π is equal to negative four. Substituting into the quadratic
formula then, we have that π₯ is equal to negative π plus or minus the square root
of π squared minus four multiplied by four multiplied by negative four all over two
multiplied by four. In the denominator, two times four
is eight. And within the square root, four
times four times negative four is negative 64. And subtracting negative 64 is the
same as adding 64. So, we have negative π plus or
minus the square root of π squared plus 64 all over eight.

Now, weβre told this key piece of
information, that the sum of the roots of this equation is negative one. Adding the expressions for our two
roots then, and we have the equation on the screen. Now, as both of these fractions
have a common denominator of eight, we can actually combine to a single
fraction. And what we notice is that thereβs
actually quite a lot of simplification. We have plus the square root of π
squared plus 64, and then minus the square root of π squared plus 64. So, these two terms will cancel
each other out. Weβre left with the far simpler
equation negative two π over eight is equal to negative one, which we can solve by
multiplying by eight and dividing by negative two to give π equals four.

In fact, this is illustrative of a
general and really useful result. When we added our two roots
together, the part under the square root canceled out due to them having different
signs. So, what we were left with, the
contribution from each root to the sum, was just negative π over two π. We therefore summed two lots of
this, giving negative two π over two π, which simplified to negative π over
π. In our case, we found that π
equals four. So, the value of π in our
quadratic equation is four. And so, negative π over π gives
negative four over four, which is negative one, the correct sum for the roots. What this has shown though is that,
in general, the sum of the roots of a quadratic equation is equal to negative π
over π. And we can quote this as a general
result.

Now that we know the value of π,
we need to determine the values of π₯. We have negative four plus or minus
the square root of four squared plus 64 over eight. We can simplify the surd here. The square root of 80 is equal to
four root five. And then dividing through by a
common factor of four gives our simplified solutions as negative one plus or minus
the square root of five over two. So, weβve completed the
problem. The value of π is four, and the
solution set of the equation in surd form is negative one minus root five over two,
negative one plus root five over two.

Letβs summarize then what weβve
seen in this video. The quadratic formula can be used
to solve quadratic equations of the form ππ₯ squared plus ππ₯ plus π is equal to
zero. The solution set or roots of this
equation, if they exist, are given by π₯ equals negative π plus or minus the square
root of π squared minus four ππ all over two π. In our final example, we also saw
that the sum of the roots of the quadratic equation ππ₯ squared plus ππ₯ plus π
equals zero is negative π over π, which we can use as a general result.