In this explainer, we will learn how to work out th roots of integers, where is a positive integer greater than or equal to 2.
The th root of a number is the inverse operation to raising a number to the th exponent, so we will first review how we find different exponents of a number.
We can recall that a number with an exponent of 1 equals the number itself; for example,
A number with an exponent of 2 would be that number multiplied by itself; for example,
Taking the square root of a number is the inverse operation to raising a number to the second exponent. We can define the square root of a number using the radical symbol .
Definition: The Square Root of a Number
A square root of a number is a number , such that . The positive square root of is written as
When we consider the square root function, , this function takes a positive input and returns the positive -value that would be squared to obtain . For example,
However, it is important to remember that we could also square the value to give 25, as
Unless otherwise directed, we usually just consider the positive value of the root. When we need to consider both the positive and negative square root values, we can indicate this by using the symbol. For example,
We can now investigate what happens when the exponent of a number is an integer greater than 2.
We could write that
And, an exponent of 4 would give
The inverse operations to both of these are the th roots, where represents the value of the exponent.
Since then the third root of 125 will be a number that is cubed to give 125, which we know is 5. We can write this as
For our value, , with an exponent of 4, since the 4th root of 125 can be found by
Note that the radical symbol is also used for th roots, and the degree of the radical symbol is the value of . When the square root is being found, we do not need to write the degree, 2, with the radical symbol. We can define the th root as follows.
Definition: πth Root of a Number
An th root, , of a quantity is a value such that
It is, therefore, the inverse function to raising a number to the th exponent. The th root is denoted
The 2nd root is usually called the square root. The 3rd root is often called the cube root. After that, the roots are simply the th root, for example, the 5th root and the 9th root.
It is often useful to make a note of some of the first few powers of 2, 3, and 4 in order to help us more quickly find the th roots of these values.
The First Five Powers of 2, 3, and 4
Learning these powers will allow for faster recall and is particularly useful for finding th roots. For example, to find , we can use the fact that to give the evaluation that .
We can now look at some example questions where we find different th roots.
Example 1: Evaluating a Square Root
Evaluate .
Answer
We can remember that when finding the square root, we do not need to write the value 2 within the square root. For example, is equivalent to . So, if we denote the root as , then which means that
So, for which positive value of will this hold true? We remember that
Therefore,
We will now see how the fourth root of a value can be evaluated.
Example 2: Evaluating a Fourth Root
Evaluate .
Answer
The th root, , of a quantity is a value such that
It is, therefore, the inverse function to raising a number to the th exponent. The th root is denoted
Here, we are asked to find ; therefore, we need to find the positive value of such that
We recall that
Therefore, , and we can write
In the following example, we will find an unknown value whose square root is equal to the cube root of another number.
Example 3: Equating Cube Roots and Square Roots
Complete the following: .
Answer
We can evaluate the left side of this equation first, . An th root, , of a quantity is a value such that
It is the inverse function to raising a number to the th exponent. The th root is denoted
In this case, the value of as we are finding the 3rd root. To find the value of , where , this is equivalent to the value such that
As we know that we can also say that
It is important to note that the missing value in the question is not simply 3. Instead, both sides of the equation must equal 3:
For the right-hand side of this equation to be equal to 3, we need to find the missing value such that
We recall that
Therefore,
Hence, the missing value is 9, and the completed equation is
In the next two examples, we will see how a number of different roots can be evaluated within one equation.
Example 4: Finding an Unknown Variable Using πth Roots
Given that , find .
Answer
In order to find the value of , we will need to evaluate the th roots within this equation.
An th root, , of a quantity is a value such that
It is the inverse function to raising a number to the th exponent. The th root is denoted
Beginning with the first term on the right side of this equation, , we can write as . Then, we find the positive value of such that . If using a trial and improvement solution, as 32 is a relatively low value, it would be sensible to begin with a low value of . We remember that any integer exponent of 1 will simply be 1; that is, ; so we could use 2 as the first trial for .
Since therefore
This means that the 5th root of 32 is 2, so we can write
We can apply the same method to find the value of the next term, . We will need to find the positive value of such that
For a trial and improvement solution, we can additionally apply a bit of logic. An even number raised to any positive integer exponent will always be even. As 625 is an odd number, this means that the value of must also be an odd number.
We can establish that
Next, we can trial the next odd value, . This gives us
Therefore,
For the final term, , we can recall that since we can write that
We can now substitute the values , , and into the equation to find the value of , noting that the term is being subtracted:
Simplifying this expression, we notice that there are terms that must be added and subtracted. We apply the order of operations to simplify this expression, giving us
We will now consider finding the th roots of negative values.
For example, is there a real value for which
We know that we can square a positive and negative value in the following example:
However, both of these give a result of 9, not . So, neither 3 nor is a solution to .
Is there a real value for which ?
If is a positive value, then would be a positive multiplied by a positive, giving a positive value. Thus , and not .
If is a negative value, then would be a negative multiplied by a negative, also giving a positive value, and not .
Thus, there is no real value of for which , and no real solution for . In general, there is no real solution for the square root of any negative number.
We can also extend this to any positive, even value of in the th root of a negative number.
We consider that and
Therefore, if we consider the 4th root of , is it possible to have a real value for which
The answer is no, as the only two potential solutions, and , both give an answer of positive 81 when raised to the 4th exponent.
Next, we can examine the situations where there are th roots of negative numbers.
For example, are there any real values of for which
We can calculate that
Therefore,
Taking a different calculation, as we can evaluate we know that
In both cases above, we found a real solution for the root of a negative number, when the roots were 3 and 5. In general, we can find a real solution for the th root of a negative number when is a positive, odd integer.
We can summarize the results in the table below.
For the equation ,
| is an even positive integer | Two real roots, | No real roots |
|---|---|---|
| is an odd positive integer | One real root, | |
We will see how to apply this in our next example.
Example 5: Finding an Unknown Variable Using πth Roots
Given that , find .
Answer
To find the value of , we will need to evaluate the th roots within this equation.
An th root, , of a quantity is a value such that
It is the inverse function to raising a number to the th exponent. The th root is denoted
Taking the terms on the left side of the equation, we can evaluate by considering which positive value of would give
As and, as such,
Since we can conclude that
We then need to evaluate . Note that when finding th roots of a value, when is an even number, we cannot find a real solution to an th root of a negative number. For example, we may recall that we cannot find a real solution for the square root of a negative number. However, when the value of is odd, we can find a real solution to an th root of a negative number.
Since we have we know that
Therefore,
The final term to evaluate is . So, which positive value of solves the equation ?
We can write that
Therefore,
We can now solve the equation by using the values , , , and . Substituting these values for the terms, and then simplifying, gives
The solution for the value of is 6.
Key Points
- The th root of a value is denoted . It is the inverse operation to raising a number to the th exponent, and we can consider this as finding the value , such that .
- We can find a real solution to the th root of a negative number when is odd.
- We cannot find a real solution to the th root of a negative number when is even.
