### Video Transcript

In this video, weβll learn how to
simplify algebraic expressions and solve algebraic equations involving πth roots,
where π is a positive integer greater than or equal to two. The πth root is a really important
mathematical operation. It describes the inverse of a power
operation for some exponent π. We define it formally by saying
that the πth root of a number π₯, where π is a positive integer, is a number that
when raised to the πth power gives π₯. In other words, if we define that
to be π¦, π₯ can be said to be equal to π¦ to the πth power. Then, the πth root is written as
shown. π¦ is equal to the πth root of
π₯.

Now, whilst itβs outside the scope
of this lesson to investigate in too much detail, it is worth noting that the πth
root of π₯ can be equivalently written as π₯ to the power of one over π. And this is really useful as it can
act as a tool that will help us understand how the rules of exponents that we
already know can be applied to expressions involving roots. The first such property tells us
what happens when we multiply them. It says that when the πth root of
π and the πth root of π are real and well defined, then the πth root of π times
π is also defined. Itβs given by the πth root of π
times the πth root of π equals the πth root of π times π.

Similarly, we can calculate the
quotient of the πth root of π over the πth root of π, where the πth root of π
is not equal to zero, by finding the πth root of π over π. Then, we consider the next four
properties very carefully. Firstly, if π is an odd integer,
then the πth root of π to the πth power is equal to the πth root of π to the
πth power, which is simply equal to π. If, however, π is even and π is
greater than or equal to zero, then the πth root of π raised to the πth power is
simply equal to π. But under the same circumstances,
if π is less than zero, then the πth root of π all raised to the πth power is
undefined over the set of real numbers.

The final property tells us what to
do if π is even and π is any real number. This time, it tells us we can take
the πth root of π to the πth power by finding the absolute value or modulus of
π. And we might be starting to wonder
what is actually happening when π is even. And weβll investigate this in a
little more detail a bit later on in the video. For now, letβs just look at a
really simple example that will allow us to use these properties to simplify an
expression involving an πth root.

Simplify the cube root of 64 times
π cubed.

In order to be able to simplify an
expression involving an πth root, where here π is equal to three, weβll recall one
of the properties that applies to πth roots. This property tells us what happens
when we multiply a pair of πth roots. Specifically, for positive real
numbers π and π, the πth root of π times the πth root of π is equivalent to
the πth root of ππ. Weβre going to apply this property
in reverse. And it allows us to separate the
cube root of 64π cubed into the product of the cube root of 64 and the cube root of
π cubed.

The next property weβre interested
in tells us that if π is an odd integer, which it is here, itβs three, then the
πth root of π all raised to the πth power is equal to the πth root of π to the
πth power, which is simply equal to π. And this is great. This allows us to simplify this
part of the expression, the cube root of π cubed. Since the root is odd, in other
words, π is equal to three, we can say that the cube root of π cubed is simply
equal to π. And of course, we know the value of
the cube root of 64. Itβs simply four. So we can substitute the cube root
of π cubed equals π and the cube root of 64 equals four back into our earlier
equation. And that will allow us to simplify
the original expression.

When we do, we find that the cube
root of 64 times the cube root of π cubed is four times π, which can of course be
fully simplified to four π. And so, weβve simplified the cube
root of 64π cubed. Itβs four π.

In this example, we looked at what
happened when the πth root, the value of π, was an odd integer. In our next example, weβll look at
finding an even root.

Simplify the square root of 100π₯
to the 16th power.

In this question, the value of π
in our πth root is omitted. When it is omitted, we assume that
itβs equal to two, and thatβs why we define it as being the square root. So weβre going to simplify this
expression by applying some of the properties of πth roots, where π is equal to
two. The first property we apply looks
at finding the product of πth roots. This tells us that when the πth
root of π and the πth root of π are well defined and real numbers, then the πth
root of π times π is also defined. Itβs the πth root of π times the
πth root of π. We can omit this value of π and
say that equivalently the square root of π times the square root of π is equal to
the square root of ππ.

We can then apply this property in
reverse. And it allows us to separate the
square root of 100π₯ to the 16th power and write it as the square root of 100 times
the square root of π₯ to the 16th power. And of course, if we know our
square numbers by heart, the first part of this expression is quite easy to
evaluate. The square root of 100 is simply
equal to 10. But what do we do with the square
root of π₯ to the 16th power? Well, weβre going to use the rule
that applies when π is an even integer and π is a real number. This is that the πth root of π to
the πth power is equal to the absolute value of π.

To be able to apply this rule, we
do need to do a little bit of manipulation, and that involves applying one of our
exponent laws. This says that π₯ to the eighth
power squared is equal to π₯ to the 16th power. So we can equivalently write the
square root of π₯ to the 16th power as the square root of π₯ to the eighth power
squared. And then, by the earlier property,
we can say that this is actually equal to the absolute value of π₯ to the eighth
power. We can now replace each part of our
earlier expression with the values 10 and the absolute value of π₯ to the eighth
power. And we see that weβve simplified
the square root of 100π₯ to the 16th power as 10 times the absolute value of π₯ to
the eighth power.

Then, letβs quickly think about
what the absolute value actually tells us. It takes the input, which is here
π₯ to the eighth power, and it makes it positive. Of course, if π₯ is a real number,
then π₯ to the πth power will actually be nonnegative for even values of π. In this case, our power eight is
even, so we can say that π₯ to the eighth power will be nonnegative. It will be greater than or equal to
zero if π₯ is a real number. This means the absolute value
symbols are actually unnecessary here, and so we can further simplify our
expression. When we do, we find that the square
root of 100π₯ to the 16th power can be simplified to 10 times π₯ to the eighth
power.

Now itβs worth noting that care
must be taken when finding powers of roots. In our previous example, we ended
up with an even power inside an even root. But what would happen if we had to
simplify something like the square root of π₯ to the power of 14? We could start in the same way by
writing π₯ to the 14th power as π₯ to the seventh power squared. And then this means that the square
root of π₯ to the 14th power is the square root of π₯ to the seventh power squared,
which is then equal to the absolute value of π₯ to the seventh power. This time, we need to keep those
absolute value symbols. This essentially ensures that no
matter the input value for π₯, the radical will be positive. This then means weβre taking the
even root of a positive number, and so itβs well defined.

Now, weβve used properties of roots
so far in this video to express an πth root as the product of two unique πth
roots. Itβs worth noting that we can
extend this to express the root as the product of three or more πth roots. Once again, if the πth root of π,
the πth root of π, and the πth root of π are well defined, then the product is
equal to the πth root of πππ, where π needs to be a positive integer.

Letβs demonstrate this in our next
example.

Write the square root of 25π
squared π to the sixth power in its simplest form.

In order to simplify this
expression, weβre going to recall that if the πth root of π, the πth root of π,
and the πth root of π are well defined for positive integers π, then their
product is the πth root of πππ. Weβre going to apply this property
in reverse to allow us to write the original expression as the square root of 25
times the square root of π squared times the square root of π to the sixth
power. Now, we know that the square root
of 25 is simply equal to five. But what do we do with the square
root of π squared and the square root of π to the sixth power?

Well, since weβre finding the
square root, this is an πth root where π is even; itβs two. So we use the following
property. If π is even and π is a real
number, then the πth root of π to the πth power is equal to the absolute value of
π. So we can actually rewrite the
square root of π squared as the absolute value of π. To repeat this process for the
square root of π to the sixth power, we need to rewrite π to the sixth power as π
cubed squared, meaning that the square root of π cubed all squared is equal to the
absolute value of π cubed.

We can now replace each root with
the value we found. So the square root of 25 times the
square root of π squared times the square root of π to the sixth power is five
times the absolute value of π times the absolute value of π cubed. Then, since five is by its very
nature nonnegative, we can use the rules for multiplying absolute value expressions
to rewrite this as the absolute value of five ππ cubed. And so in its simplest form, the
square root of 25π squared π to the sixth power is equal to the absolute value of
five ππ cubed.

So weβve now demonstrated how to
use properties of roots to simplify expressions. But itβs worth being aware that we
do need to be careful when working with equations that involve exponents. Consider, for instance, the
equation π¦ squared equals 16. We would look to solve this
equation by taking the square root of both sides. So a solution to the equation is π¦
equals the square root of 16, which is four. However, if we substitute π¦ equals
negative four into the expression π¦ squared, we get negative four squared, which is
in fact positive 16. So thereβs a second solution to
this equation. Itβs negative four. And so when solving an equation of
this form, the solutions need to involve both the positive and negative square roots
of π₯. And so thereβs a real subtle
difference between the two seemingly equivalent statements, π¦ squared equals π₯ and
π¦ equals the square root of π₯.

Letβs look to generalize this. Letβs consider the equation π¦ to
the πth power equals π₯, where π₯ and π¦ are real numbers and π is a positive
integer. Firstly, if π is an even number,
but π₯ is negative, then there are no real solutions to the equation π¦ to the πth
power equals π₯. If, however, π is even and π₯ is
positive, the solutions are given by π¦ equals plus or minus the πth root of
π₯. Of course, if π₯ is equal to zero,
then the solution is just π¦ equals zero. If, however, π is odd, it doesnβt
matter the sign of π₯. There will always be one solution
to the equation, and that is π¦ is equal to the πth root of π₯.

Now, because we interpret the
statements π¦ to the πth power equals π₯ and π¦ is equal to the πth root of π₯
differently, we do have one further definition, and thatβs of the principal πth
root. With this definition, we are able
to consider an πth root as a function by making it by definition a one-to-one
mapping. We say that every positive real
number has a single positive πth root; thatβs the πth root of π₯. This is known as the principal πth
root. So when given, for instance, the
expression the square root of four, we know we are only interested in two and not
negative two because weβre not solving an equation.

Letβs demonstrate an application of
the properties of even and odd πth roots in our next example.

Find the value, or values, of π₯ if
12π₯ to the fifth power equals 384.

Weβre going to solve this equation
for π₯ by performing a series of inverse operations. We notice that π₯ to the fifth
power is being multiplied by 12. So weβre going to begin by dividing
both sides of this equation by 12. 384 divided by 12 is 32. So we get π₯ to the fifth power
equals 32. Then, to solve this equation, we
recall what we know about even and odd roots. Specifically, given the equation π¦
to the πth power equals π₯, if π is odd, then there is exactly one solution to
this equation, π¦ equals the πth root of π₯.

Now, since our equation is in terms
of π₯, weβre going to rewrite this slightly. So if π₯ to the πth power equals
π¦, then π₯ is equal to the πth root of π¦. Now, in this case, π is odd; itβs
equal to five. And so there is just one solution
to this equation. Itβs π₯ equals the fifth root of
32. But of course, the fifth root of 32
is simply equal to two. And so, the solution is π₯ equals
two.

Weβre going to consider just one
more example. This time, weβre going to combine
the properties of πth roots and the theorem about even and odd roots we just
demonstrated. This will allow us to solve more
complicated equations involving exponents.

Find the value, or values, of π₯
given that π₯ plus nine over five squared is equal to the square root of 144 times
three squared.

Weβre going to begin by evaluating
the right-hand side of this equation. We know we can write it as the
square root of 144 times the square root of three squared. Then, we apply the property that if
π is even and π is a real number, then the πth root of π to the πth power is
equal to the absolute value of π. Specifically, the second part of
this expression can be thought of as the square root of three squared. And so we can rewrite that as the
absolute value of three. Now, of course, the square root of
144 is 12, but three is nonnegative. So we donβt actually need the
absolute value symbols. This gives us 12 times three, which
is equal to 36. So letβs rewrite our equation as
shown.

Next, we think about what we know
about solving equations involving odd and even roots. Specifically, if we have even π
and positive π₯, the solutions to the equation π¦ to the πth power equals π₯ are π₯
equals positive or negative the πth root of π¦. This means if we take the square
root of both sides of this equation here, we have to find the positive and negative
square root of 36. And in fact, the square root of 36
is six. So our equation can be written as
π₯ plus nine over five equals positive or negative six.

And so to find the values of π₯
that satisfy our original equation, weβre going to solve two equations: π₯ plus nine
over five equals six and π₯ plus nine over five equals negative six. We begin by multiplying each
equation by five. So our first equation can then be
written as π₯ plus nine equals 30 and our second as π₯ plus nine equals negative
30. Then, we subtract nine, and we get
π₯ equals 21 or π₯ equals negative 39. And so we found the values of π₯
that satisfy our equation. And if we were to substitute either
of these back into this equation, this would work. We get π₯ equals 21 and π₯ equals
negative 39.

Letβs now recap the key points from
this lesson. In this lesson, we learned that the
πth root of π₯, where π is a positive integer, is some number π¦ such that π₯ is
equal to π¦ to the πth power. And itβs written as shown. We learned the definition of the
principal root, and we said that every positive real number has a single πth root
called the principal root. We learned how to extend the laws
of exponents to working with πth roots. And we learned how to find the πth
power of πth roots and vice versa when π is odd. We saw that there are three
separate scenarios we should consider if π is even. And we saw that we should take
great care when solving equations of the form π¦ to the πth power equals π₯.