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Lesson Video: 𝑛th Roots: Expressions and Equations Mathematics

In this video, we will learn how to simplify algebraic expressions and solve algebraic equations involving 𝑛th roots, where 𝑛 is a positive integer greater than or equal to 2.

18:00

Video Transcript

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

Now, whilst it’s outside the scope of this lesson to investigate in too much detail, it is worth noting that the 𝑛th root of π‘₯ can be equivalently written as π‘₯ to the power of one over 𝑛. 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 𝑛th root of π‘Ž and the 𝑛th root of 𝑏 are real and well defined, then the 𝑛th root of π‘Ž times 𝑏 is also defined. It’s given by the 𝑛th root of π‘Ž times the 𝑛th root of 𝑏 equals the 𝑛th root of π‘Ž times 𝑏.

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

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

Simplify the cube root of 64 times π‘š cubed.

In order to be able to simplify an expression involving an 𝑛th root, where here 𝑛 is equal to three, we’ll recall one of the properties that applies to 𝑛th roots. This property tells us what happens when we multiply a pair of 𝑛th roots. Specifically, for positive real numbers π‘Ž and 𝑏, the 𝑛th root of π‘Ž times the 𝑛th root of 𝑏 is equivalent to the 𝑛th root of π‘Žπ‘. We’re going to apply this property in reverse. And it allows us to separate the cube root of 64π‘š cubed into the product of the cube root of 64 and the cube root of π‘š cubed.

The next property we’re interested in tells us that if 𝑛 is an odd integer, which it is here, it’s three, then the 𝑛th root of π‘Ž all raised to the 𝑛th power is equal to the 𝑛th root of π‘Ž to the 𝑛th power, which is simply equal to π‘Ž. And this is great. This allows us to simplify this part of the expression, the cube root of π‘š cubed. Since the root is odd, in other words, 𝑛 is equal to three, we can say that the cube root of π‘š cubed is simply equal to π‘š. And of course, we know the value of the cube root of 64. It’s simply four. So we can substitute the cube root of π‘š cubed equals π‘š 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‘š cubed is four times π‘š, which can of course be fully simplified to four π‘š. And so, we’ve simplified the cube root of 64π‘š cubed. It’s four π‘š.

In this example, we looked at what happened when the 𝑛th root, the value of 𝑛, was an odd integer. In our next example, we’ll look at finding an even root.

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

In this question, the value of 𝑛 in our 𝑛th root is omitted. When it is omitted, we assume that it’s equal to two, and that’s why we define it as being the square root. So we’re going to simplify this expression by applying some of the properties of 𝑛th roots, where 𝑛 is equal to two. The first property we apply looks at finding the product of 𝑛th roots. This tells us that when the 𝑛th root of π‘Ž and the 𝑛th root of 𝑏 are well defined and real numbers, then the 𝑛th root of π‘Ž times 𝑏 is also defined. It’s the 𝑛th root of π‘Ž times the 𝑛th root of 𝑏. We can omit this value of 𝑛 and say that equivalently the square root of π‘Ž times the square root of 𝑏 is equal to the square root of π‘Žπ‘.

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, we’re going to use the rule that applies when 𝑛 is an even integer and π‘Ž is a real number. This is that the 𝑛th root of π‘Ž to the 𝑛th power is equal to the absolute value of π‘Ž.

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 we’ve simplified the square root of 100π‘₯ to the 16th power as 10 times the absolute value of π‘₯ to the eighth power.

Then, let’s 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 𝑛th power will actually be nonnegative for even values of 𝑛. 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 it’s 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 we’re taking the even root of a positive number, and so it’s well defined.

Now, we’ve used properties of roots so far in this video to express an 𝑛th root as the product of two unique 𝑛th roots. It’s worth noting that we can extend this to express the root as the product of three or more 𝑛th roots. Once again, if the 𝑛th root of π‘Ž, the 𝑛th root of 𝑏, and the 𝑛th root of 𝑐 are well defined, then the product is equal to the 𝑛th root of π‘Žπ‘π‘, where 𝑛 needs to be a positive integer.

Let’s demonstrate this in our next example.

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

In order to simplify this expression, we’re going to recall that if the 𝑛th root of π‘Ž, the 𝑛th root of 𝑏, and the 𝑛th root of 𝑐 are well defined for positive integers 𝑛, then their product is the 𝑛th root of π‘Žπ‘π‘. We’re 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‘Ž squared times the square root of 𝑏 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‘Ž squared and the square root of 𝑏 to the sixth power?

Well, since we’re finding the square root, this is an 𝑛th root where 𝑛 is even; it’s two. So we use the following property. If 𝑛 is even and π‘Ž is a real number, then the 𝑛th root of π‘Ž to the 𝑛th power is equal to the absolute value of π‘Ž. So we can actually rewrite the square root of π‘Ž squared as the absolute value of π‘Ž. To repeat this process for the square root of 𝑏 to the sixth power, we need to rewrite 𝑏 to the sixth power as 𝑏 cubed squared, meaning that the square root of 𝑏 cubed all squared is equal to the absolute value of 𝑏 cubed.

We can now replace each root with the value we found. So the square root of 25 times the square root of π‘Ž squared times the square root of 𝑏 to the sixth power is five times the absolute value of π‘Ž times the absolute value of 𝑏 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 π‘Žπ‘ cubed. And so in its simplest form, the square root of 25π‘Ž squared 𝑏 to the sixth power is equal to the absolute value of five π‘Žπ‘ cubed.

So we’ve now demonstrated how to use properties of roots to simplify expressions. But it’s worth being aware that we do need to be careful when working with equations that involve exponents. Consider, for instance, the equation 𝑦 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 𝑦 equals the square root of 16, which is four. However, if we substitute 𝑦 equals negative four into the expression 𝑦 squared, we get negative four squared, which is in fact positive 16. So there’s a second solution to this equation. It’s 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 there’s a real subtle difference between the two seemingly equivalent statements, 𝑦 squared equals π‘₯ and 𝑦 equals the square root of π‘₯.

Let’s look to generalize this. Let’s consider the equation 𝑦 to the 𝑛th power equals π‘₯, where π‘₯ and 𝑦 are real numbers and 𝑛 is a positive integer. Firstly, if 𝑛 is an even number, but π‘₯ is negative, then there are no real solutions to the equation 𝑦 to the 𝑛th power equals π‘₯. If, however, 𝑛 is even and π‘₯ is positive, the solutions are given by 𝑦 equals plus or minus the 𝑛th root of π‘₯. Of course, if π‘₯ is equal to zero, then the solution is just 𝑦 equals zero. If, however, 𝑛 is odd, it doesn’t matter the sign of π‘₯. There will always be one solution to the equation, and that is 𝑦 is equal to the 𝑛th root of π‘₯.

Now, because we interpret the statements 𝑦 to the 𝑛th power equals π‘₯ and 𝑦 is equal to the 𝑛th root of π‘₯ differently, we do have one further definition, and that’s of the principal 𝑛th root. With this definition, we are able to consider an 𝑛th 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 𝑛th root; that’s the 𝑛th root of π‘₯. This is known as the principal 𝑛th 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 we’re not solving an equation.

Let’s demonstrate an application of the properties of even and odd 𝑛th roots in our next example.

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

We’re 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 we’re 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 𝑦 to the 𝑛th power equals π‘₯, if 𝑛 is odd, then there is exactly one solution to this equation, 𝑦 equals the 𝑛th root of π‘₯.

Now, since our equation is in terms of π‘₯, we’re going to rewrite this slightly. So if π‘₯ to the 𝑛th power equals 𝑦, then π‘₯ is equal to the 𝑛th root of 𝑦. Now, in this case, 𝑛 is odd; it’s equal to five. And so there is just one solution to this equation. It’s 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.

We’re going to consider just one more example. This time, we’re going to combine the properties of 𝑛th 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.

We’re 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 𝑛 is even and π‘Ž is a real number, then the 𝑛th root of π‘Ž to the 𝑛th power is equal to the absolute value of π‘Ž. 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 don’t actually need the absolute value symbols. This gives us 12 times three, which is equal to 36. So let’s 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 𝑛 and positive π‘₯, the solutions to the equation 𝑦 to the 𝑛th power equals π‘₯ are π‘₯ equals positive or negative the 𝑛th root of 𝑦. 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, we’re 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.

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

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