### Video Transcript

In this video, we will learn how to
solve quadratic equations using the quadratic formula. The quadratic formula in its
current form was first recorded in the 17th century by the French mathematician Rene
Descartes. It is an efficient way to solve a
quadratic equation, especially one which cannot be solved by factoring. Given any quadratic equation in the
form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants
and π is nonzero, then the quadratic formula states that π₯ is equal to negative π
plus or minus the square root of π squared minus four ππ all divided by two
π. Letβs now consider how we can use
this formula to solve any problem involving a quadratic equation.

Our first step is to write our
equation in the form ππ₯ squared plus ππ₯ plus π equals zero. In order to do this, we might need
to distribute parentheses or brackets as well as collect like terms. Whilst it is not essential, it is
useful to write our equation in the same order. It is then easy to list the values
of π, π, and π. Next, we substitute our values of
π, π, and π into the quadratic formula. From this point, we simplify the
right-hand side obtaining two solutions to the quadratic equation. Letβs consider the quadratic
equation three π₯ squared plus four π₯ minus one is equal to zero. Our values of π, π, and π are
three, four, and negative one, respectively, as the coefficient of π₯ squared is
three, the coefficient of π₯ is four, and the term independent of π₯ is negative
one.

Substituting these values into the
quadratic formula gives us π₯ is equal to negative four plus or minus the square
root of four squared minus four multiplied by three multiplied by negative one all
divided by two multiplied by three. Underneath the square root, our
expression simplifies to 16 plus 12. This gives us negative four plus or
minus the square root of 28 all divided by six. The square root of 28 is equal to
two root seven. We can then divide each term in our
expression by two. π₯ is therefore equal to negative
two plus or minus root seven all divided by three. We can separate this to get the two
solutions to the quadratic equation three π₯ squared plus four π₯ minus one equals
zero. Either π₯ is equal to negative two
plus root seven divided by three or π₯ is equal to negative two minus root seven
divided by three.

We will now look at a more
complicated example.

Find the solution set of the
equation three π₯ squared plus four multiplied by π₯ plus one equals zero, giving
values in the set of real numbers to one decimal place.

In order to answer this question,
we firstly need to rearrange our equation so it is in the form ππ₯ squared plus
ππ₯ plus π equals zero. This will enable us to use the
quadratic formula to solve it. Distributing the parentheses by
multiplying four by π₯ and four by one gives us three π₯ squared plus four π₯ plus
four equals zero. Our values of π, π, and π are
three, four, and four, respectively. The quadratic formula states that
π₯ is equal to negative π plus or minus the square root of π squared minus four
ππ all divided by two π.

Substituting in our values, we have
π₯ is equal to negative four plus or minus the square root of four squared minus
four multiplied by three multiplied by four all divided by two multiplied by
three. Four squared is 16, four multiplied
by three multiplied by four is 48, and two multiplied by three is six. As 16 minus 48 is negative 32, we
are left with π₯ is equal to negative four plus or minus the square root of negative
32 all divided by six.

At this stage, as we want solutions
to one decimal place, we would usually input our calculation into our
calculator. However, this calculation involves
taking the square root of a negative number, negative 32. And we know that square rooting a
negative number has no real solutions. And when we type it into the
calculator, we get a mathematical error. This means that there are no real
solutions to our equation. And the solution set of the
equation is the empty set. This leads us to a key fact about
the quadratic formula. If the expression underneath the
square root π squared minus four ππ, known as the discriminant, is less than
zero, then there are no real solutions to our quadratic equation.

It is also worth considering what
this would look like graphically. The quadratic equation π¦ is equal
to three π₯ squared plus four multiplied by π₯ plus one is shown in the figure. We notice that the graph does not
intersect the π₯-axis. This confirms that there are no
real solutions to the equation. Any quadratic equation, where the
discriminant π squared minus four ππ is less than zero, will not intersect the
π₯-axis.

We will now look at two further
examples where we need to use the quadratic formula in context.

The dimensions of a rectangle are
five meters and 12 meters. When both dimensions are increased
by a given amount, the area of the rectangle will double. What is the amount?

We are given that the dimensions of
a rectangle are five meters and 12 meters. Recalling that we can calculate the
area of a rectangle by multiplying its length by its width, the area of this
rectangle is 60 square meters. Next, we are told that both
dimensions are increased by a given amount. We will let this given amount be π₯
meters so that the length of the new rectangle is 12 plus π₯ meters and the width is
five plus π₯ meters.

To calculate the area of this
rectangle, we need to multiply 12 plus π₯ and five plus π₯. We are also told that the area of
the rectangle has doubled. Therefore, the area of this
rectangle is 120 square meters. Distributing our parentheses using
the FOIL method, we have 60 plus 12π₯ plus five π₯ plus π₯ squared. And this is equal to 120. Subtracting 120 from both sides of
this equation and then collecting like terms, we have π₯ squared plus 17π₯ minus 60
is equal to zero. This is a quadratic equation
written in the form ππ₯ squared plus ππ₯ plus π equals zero. And we know we can solve a
quadratic equation of this type using the quadratic formula. This states that π₯ is equal to
negative π plus or minus the square root of π squared minus four ππ all divided
by two π.

In this question, our values of π,
π, and π are one, 17, and negative 60, respectively. Substituting these values into the
formula, we have π₯ is equal to negative 17 plus or minus the square root of 17
squared minus four multiplied by one multiplied by negative 60 all divided by two
multiplied by one. The expression under the square
root is equal to 529. And as the square root of 529 is
23, we have π₯ is equal to negative 17 plus or minus 23 all divided by two. This gives us two possible
solutions, either π₯ is equal to negative 17 plus 23 divided by two or π₯ is equal
to negative 17 minus 23 all divided by two. These are equal to three and
negative 20, respectively.

Since we are solving to find the
dimension π₯ of the rectangle, we know that π₯ must be positive. This means that the correct
solution is π₯ equals three. For the area of the rectangle to
double, both dimensions need to be increased by 3 meters. This would give us a rectangle with
dimensions eight meters by 15 meters. And multiplying these gives us the
correct area of 120 square meters. It is important to note that this
quadratic equation could have been factored. Using this method would also have
given us solutions of π₯ equals negative 20 and π₯ equals three.

We will now consider one final
example.

A stone is thrown upward from the
top of a cliff and lands in the sea some time later. The height β above sea level at any
time π‘ seconds is given by the formula β is equal to three π‘ minus five π‘ squared
plus 40. After how many seconds will the
stone reach the sea? Give your answer to the nearest
tenth of a second.

In this question, we are given a
quadratic equation for β in terms of π‘. Rewriting the right-hand side in
the form ππ₯ squared plus ππ₯ plus π, we have β is equal to negative five π‘
squared plus three π‘ plus 40. We are trying to calculate the time
at which the stone will reach the sea, and we are told that β is the height above
sea level. This means that when the stone
reaches the sea, β will be equal to zero. And we need to solve the quadratic
equation negative five π‘ squared plus three π‘ plus 40 equals zero. We know that any quadratic equation
ππ₯ squared plus ππ₯ plus π equals zero can be solved using the quadratic
formula. This states that π₯ is equal to
negative π plus or minus the square root of π squared minus four ππ all divided
by two π.

In this question, our values of π,
π, and π are negative five, three, and 40, and the variable instead of being π₯ is
the time π‘. Substituting in our values, we have
π‘ is equal to negative three plus or minus the square root of three squared minus
four multiplied by negative five multiplied by 40 all divided by two multiplied by
negative five. This simplifies to negative three
plus or minus the square root of 809 all divided by negative 10. Separating our two solutions, we
have π‘ is equal to negative three plus the square root of 809 divided by negative
10 or π‘ is equal to negative three minus the square root of 809 divided by negative
10.

Typing these into the calculator,
we have π‘ is equal to negative 2.544 and so on and π‘ is equal to 3.144 and so
on. As the time in seconds must be a
positive value, we can rule out the first answer. We are also asked to give the time
to the nearest tenth of a second. We can therefore conclude that it
takes 3.1 seconds for the stone to reach the sea.

We will now summarize the key
points from this video. For any quadratic equation in the
form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants
and π is nonzero, we can use the quadratic formula to solve for π₯. This states that π₯ is equal to
negative π plus or minus the square root of π squared minus four ππ all divided
by two π. The solutions to the equation are
the points where the graph of the equation crosses the π₯-axis. If the discriminant π squared
minus four ππ is less than zero, then the quadratic equation has no real
solutions. This means that the graph would not
pass through the π₯-axis. Whilst it is not included in this
video, it may be of interest to research how the quadratic formula can be
derived.