### Video Transcript

In this video, we will learn how to
solve word problems by forming and solving quadratic equations. These are equations which can be
written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero for some
constant 𝑎, 𝑏, and 𝑐, which just means values, usually numbers. It must also be the case that the
value of 𝑎 isn’t zero, as this would leave us with no 𝑥 squared term. And so, the equation would be
linear not quadratic. It’s perfectly fine for 𝑏 or 𝑐 to
be zero there.

You should already be familiar with
solving quadratic equations by factorizing. You may also be familiar with two
other methods for solving quadratic equations, using the quadratic formula and
completing the square. But don’t worry if you haven’t met
these methods yet, as they won’t be required for the examples covered in this
lesson. The focus of this video will mainly
be on setting up the quadratic equation from a wordy description or diagram. But we will also solve each of our
equations. We’ll review the method of solving
a quadratic equation by factorizing or factoring in more detail in the context of
examples.

The sum of the squares of two
positive real numbers is 542. Given that one of them is 18, find
the value of the other one.

So in this question, we’re not
looking to take a trial-and-error approach. We’re going to answer this question
by forming an equation. Let’s look at the information we’ve
been given. We’re told that the sum of the
squares of two positive real numbers is 542. We need to introduce letters to
represent these numbers. So at this point you may think,
let’s call one number 𝑥 and the other 𝑦. And if the sum of their squares is
542, then we can express this as an equation, 𝑥 squared plus 𝑦 squared equals
542. However, if we read on a bit in the
question, we see that we’re told that one of the numbers is 18. So we don’t actually need to use
two letters at all. We can use one letter, 𝑥, to
represent the unknown number and then our second number is 18. So we have the equation 𝑥 squared
plus 18 squared equals 542.

Now, what we’ve done here is form
an equation. And it is a quadratic equation
because the highest power of our variable, that’s 𝑥, that appears is two. Let’s now see how we go about
solving this equation. Firstly, on the left-hand side, we
can evaluate 18 squared; it’s equal to 324. We want to get 𝑥 on its own on the
left-hand side. So the next step is to subtract 324
from each side. When we do, we’re just left with 𝑥
squared on the left-hand side. And on the right-hand side, 542
minus 324 is 218.

Now, we have 𝑥 squared equals
218. And in order to work out the value
of 𝑥, we need to apply the inverse or opposite operation of squaring, which is
square rooting. So we take the square root of each
side of the equation. The square root of 𝑥 squared is
𝑥. And on the right-hand side, we take
the square root of 218. But it’s important that when we
solve an equation by square rooting, we remember to take plus or minus the square
root. We have then that 𝑥 is equal to
plus or minus the square root of 218.

Now, this is certainly the correct
solution to the quadratic equation we’ve formed. But if we look back at the
question, we were told that both of our numbers are positive real numbers. Which means that in terms of the
answer to the question, 𝑥 must take only a positive value. So whilst negative the square root
of 218 is a valid solution to the quadratic equation, it isn’t a valid solution to
this problem. So our only answer is 𝑥 equals the
square root of 218. Now, this surd can’t be simplified
any further, as 218 has no square factors other than one. And it makes sense to give an exact
answer, so leaving our answer in terms of a square root rather than a rounded
decimal. Our answer then is that the second
number is the square root of 218.

We can, of course, check this by
substituting our value of 𝑥 back into our equation or the information given in the
question and confirming that we do indeed get 542 for the sum of the squares of the
two numbers.

Now, in this problem, the quadratic
equation we formed could be solved by simply rearranging and then square
rooting. In our next problem, we’ll look at
an example where we need to solve a quadratic equation using the method of
factorizing or factoring as you may know it.

What is the perimeter of a
rectangle whose length is seven centimeters more than its width and whose area is 78
centimeters squared?

Now, in a problem like this, it’s
always a good idea to draw our own diagram so that we can visualize the
situation. We have a rectangle, so it looks a
little something like this. Now, we don’t know its width, so we
can use the letter 𝑤 to represent this unknown value. We’re told that the length of the
rectangle is seven centimeters more than the width, which means we can express the
length in terms of the width. If the width is 𝑤, then the length
will be 𝑤 plus seven. So we have expressions for both the
length and the width of the rectangle in terms of the same letter, 𝑤. We’re also told that the area of
this rectangle is 78 centimeters squared, and we’re asked to find its perimeter.

In order to do this, we’re going to
need to know the value of 𝑤. So we can use the information we’ve
been given to form an equation. The area of a rectangle is found by
multiplying its length by its width. So, using the expressions we have
for these two values, we have 𝑤 multiplied by 𝑤 plus seven or 𝑤 plus seven
multiplied by 𝑤. But as we know, the area is 78
centimeters squared. We can write this as an
equation. We have then 𝑤 multiplied by 𝑤
plus seven equals 78.

The next step is to rearrange this
equation slightly, which we’ll do by expanding the brackets or distributing the
parentheses as you may call it. 𝑤 multiplied by 𝑤 gives 𝑤
squared, and 𝑤 multiplied by seven gives seven 𝑤. So we have 𝑤 squared plus seven 𝑤
equals 78. The next step is just subtract 78
from each side of the equation, as we want to group all of the terms together on the
same side. Doing so gives 𝑤 squared plus
seven 𝑤 minus 78 is equal to zero. And now, we see that we have a
quadratic equation in 𝑤 in this most easily recognizable form. We need to solve this equation to
find the value of 𝑤. So let’s see whether this equation
can be factored.

As the coefficient of 𝑤 squared,
that’s the number in front of 𝑤 squared, is just one, the first term in each of our
factors will just be 𝑤. And to complete our factors, we’re
then looking for two numbers whose sum is the coefficient of 𝑤, that’s positive
seven, and whose product is the constant term, that’s negative 78. To find these numbers, we can begin
by listing all the factor pairs of 78. They are one and 78, two and 39,
three and 26, and six and 13. Now, the product needs to be
negative 78, which means one of our numbers needs to be negative and the other needs
to be positive. And we see that if we use our last
factor pair and we choose six to be negative but 13 to be positive. Then we have negative six plus 13
or 13 minus six, which is indeed equal to seven. So the sum of this factor pair
would be the correct value.

So we can complete our two sets of
parentheses by adding the value negative six to the first and positive 13 to the
second. We now have the factored form of
our quadratic. And you can confirm that this is
indeed correct by redistributing or reexpanding the brackets. The next step in this method is to
recall that if the product of two factors is zero, then at least one of those
factors must themselves be zero. So we take each factor in turn and
set it equal to zero, giving 𝑤 minus six equals zero or 𝑤 plus 13 equals zero. We can then solve the resulting
linear equations.

To solve the first equation, we add
six to both sides giving 𝑤 equals six. And to solve the second, we
subtract 13 from each side giving 𝑤 equals negative 13. So we have two solutions to our
quadratic equation. 𝑤 equals six and 𝑤 equals
negative 13. However, whilst these are both
valid solutions to the quadratic equation, they aren’t both valid in terms of the
context of this problem. If we look back at our rectangle,
we can see that 𝑤 represents its width. And the width must take a positive
value. 𝑤 therefore cannot be equal to
negative 13 in this problem. So we eliminate this value. We’ve found then that the value of
𝑤 and the width of our rectangle is six.

Now, finally, we recall that we
weren’t just asked to find the width of the rectangle, but we were asked to find its
perimeter. So we now know that its width is
six centimeters and its length is six plus seven; that’s 13 centimeters. The perimeter is the distance all
the way around the edge of this rectangle. So we can either just add up all
four sides, 13 plus six plus 13 plus six. Or we could use the formula that
the perimeter is equal to twice the sum of the length and the width. So we could have two times six plus
13. In either case, it will, of course,
give the same value of 38. The area of this rectangle was
given in centimeters squared, so the units for its perimeter will be
centimeters. And we’ve found then that the
perimeter of the rectangle is 38 centimeters.

To check our answer, we can
multiply the length and the width that we’ve calculated together and confirm that
the area is indeed equal to 78 centimeters squared.

Now, in this example, the quadratic
equation we had to factor had a leading coefficient, that’s the coefficient of 𝑤
squared, equal to one. In our final example, we’ll remind
ourselves how to solve quadratic equations by factoring when the leading coefficient
is not equal to one.

Find the positive number which is
66 less than twice its square.

To answer this question, we’re
going to need to use some algebra, and we’re going to need to form an equation. So let’s introduce a letter to
represent this positive number. We can call it whatever we like;
let’s call it 𝑥. We’re told that 𝑥 is 66 less than
twice its square. Now, the square of 𝑥 can be
written as 𝑥 squared. Twice its square means we’re
multiplying this by two, so that gives two 𝑥 squared. And we then want 66 less than this
value. So we have the expression two 𝑥
squared minus 66. But remember, this is equal to the
number itself. So as an equation, we have two 𝑥
squared minus 66 is equal to 𝑥. We’ve therefore formulated this
problem as a quadratic equation. And we now need to solve it in
order to find the value of 𝑥.

We want to collect all of the terms
on the same side of the equation. Which we can do by subtracting 𝑥
from each side, giving two 𝑥 squared minus 𝑥 minus 66 equals zero, which is a
quadratic equation in its most easily recognizable form. Now, we want to solve this
quadratic equation by factoring. But notice that the coefficient of
𝑥 squared, the leading coefficient, is not equal to one. So we’re going to use the method of
factoring by grouping. We look for two numbers whose sum
is the coefficient of 𝑥, that’s negative one, and whose product is the product of
the coefficient of 𝑥 squared and the constant term. That’s two times negative 66 which
is negative 132.

The more familiar you are with your
times tables, the more quickly you’ll spot these two numbers. They’re negative 12 and positive
11. Now, the next step may look a
little strange. What we’re going to do is take that
term of negative 𝑥 and rewrite it using these two numbers. We’re going to rewrite it as
negative 12𝑥 plus 11𝑥. So we now have our quadratic
equation with four terms, two 𝑥 squared minus 12𝑥 plus 11𝑥 minus 66 is equal to
zero. Next, what we’re going to do is
split this quadratic equation in half. And we’re going to factor the two
halves separately.

Looking at the first half, two 𝑥
squared minus 12𝑥, these two terms have a common factor of two 𝑥. And if we take this out, we’re then
left with 𝑥 minus six. So the first half factors as two 𝑥
multiplied by 𝑥 minus six. Looking at the second half of our
quadratic, 11𝑥 minus 66, these two terms have a common factor of 11. And once we’ve factored by 11,
we’re left with 𝑥 minus six inside the parentheses. So the second half of our quadratic
factorizes as 11 multiplied by 𝑥 minus six.

Now, the key point, and this will
always be the case if a quadratic equation can be solved by factoring, is that the
two halves of our expression now have a common factor of 𝑥 minus six. We therefore factor the entire
quadratic by 𝑥 minus six. For the first term, we have to
multiply by two 𝑥. And for the second, we have to
multiply by positive 11. So our quadratic can be written as
𝑥 minus six multiplied by two 𝑥 plus 11. And it’s now in a fully factored
form. Remember, it will always be the
case that the two halves of your expression share a common factor if the quadratic
equation can be factored.

If you go through this method and
you find the two halves don’t share a common factor, then you’ve either made a
mistake or the quadratic equation you’re working with can’t actually be
factored. And you need to use another method
to solve it such as the quadratic formula or completing the square, if you’re aware
of these. Anyway, our quadratic equation can
be factored, and we have its factored form. So the next step is to take each
factor in turn and set them equal to zero.

We have 𝑥 minus six equals zero or
two 𝑥 plus 11 equals zero. To solve the first equation, we
need to add six to each side, giving 𝑥 equals six. And we can solve the second
equation in two steps. First, we subtract 11 from each
side, giving two 𝑥 equals negative 11. And then, we can divide by two,
giving 𝑥 equals negative 11 over two or negative 5.5. Now, both of these are valid
solutions to our quadratic equation. But if we look back at the
question, we were told that this number that we’re looking for must be positive. Which means it can’t be equal to
negative 11 over two. So we can eliminate this value. Our solution, then, is that 𝑥 is
equal to six. But let’s check this.

This number needs to be 66 less
than twice its square, so we have two times six squared minus 66. That’s two times 36 which is 72
minus 66 which is indeed equal to six, the number itself. So this confirms that our solution
is correct. So we have our answer to the
problem. The positive number we’re looking
for is six.

Let’s now summarize what we’ve seen
in this video. Firstly, we’ve seen that quadratic
equations can be used to solve many different types of problems. Including number problems and
problems involving the area of two-dimensional shapes or perhaps the surface area of
a three-dimensional shape. If we’re given a wordy description
of a physical problem, it’s often a good idea to draw a simple diagram if one hasn’t
been provided for us.

We may need to introduce letters or
variables ourselves to represent any missing values in the question. And we then need to form the
quadratic equation using the information given. We must make sure we read it
carefully and break the information down. We then need to solve our quadratic
equation by factoring. Or perhaps in the future, we could
use another method, such as the quadratic formula or completing the square. Being able to confidently form and
solve quadratic equations is a really useful skill because they can be used to
represent such a wide variety of problems.