Lesson Explainer: nth Roots: Integers Mathematics

In this explainer, we will learn how to work out 𝑛th roots of integers, where 𝑛 is a positive integer greater than or equal to 2.

The 𝑛th root of a number is the inverse operation to raising a number to the 𝑛th 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, 5=5.

A number with an exponent of 2 would be that number multiplied by itself; for example, 5=5Γ—5=25.

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 ο€Ίβˆšο†.

Definition: The Square Root of a Number

A square root of a number π‘₯ is a number 𝑦, such that 𝑦=π‘₯. The positive square root of π‘₯ is written as √π‘₯=𝑦.

When we consider the square root function, √π‘₯=𝑦, this function takes a positive input π‘₯ and returns the positive 𝑦-value that would be squared to obtain π‘₯. For example, √25=5.

However, it is important to remember that we could also square the value βˆ’5 to give 25, as βˆ’5Γ—βˆ’5=25.

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, ±√25=±5.

We can now investigate what happens when the exponent of a number is an integer greater than 2.

We could write that 5=5Γ—5Γ—5=125.

And, an exponent of 4 would give 5=5Γ—5Γ—5Γ—5=625.οŠͺ

The inverse operations to both of these are the 𝑛th roots, where 𝑛 represents the value of the exponent.

Since 5=125, 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 √125=5.

For our value, 5οŠͺ, with an exponent of 4, since 5=125,οŠͺ the 4th root of 125 can be found by √125=5.

Note that the radical symbol ο€Ίβˆšο† is also used for 𝑛th roots, and the degree of the radical symbol is the value of 𝑛. When the square root is being found, we do not need to write the degree, 2, with the radical symbol. We can define the 𝑛th root as follows.

Definition: 𝑛th Root of a Number

An 𝑛th root, π‘Ÿ, of a quantity 𝑧 is a value such that 𝑧=π‘Ÿ.

It is, therefore, the inverse function to raising a number to the 𝑛th exponent. The 𝑛th root is denoted π‘Ÿ=βˆšπ‘§.

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 𝑛th root, for example, the 5th root and the 9th root.

It is often useful to make a note of some of the first few exponents of 2, 3, and 4 in order to help us more quickly find the 𝑛th roots of these values.

The First Five Exponents of 2, 3, and 4

2=23=34=42=43=94=162=83=274=642=163=814=2562=323=2434=1024οŠͺοŠͺοŠͺ

Learning these powers will allow for faster recall and is particularly useful for finding 𝑛th roots. For example, to find √243, we can use the fact that 3=243 to give the evaluation that √243=3.

We can now look at some example questions where we find different 𝑛th roots.

Example 1: Evaluating a Square Root

Evaluate √4.

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, √4 is equivalent to √4. So, if we denote the root as π‘Ÿ, then √4=π‘Ÿ, which means that π‘ŸΓ—π‘Ÿ=π‘Ÿ=4.

So, for which positive value of π‘Ÿ will this hold true? We remember that 2Γ—2=4.

Therefore, √4=2.

We will now see how the fourth root of a value can be evaluated.

Example 2: Evaluating a Fourth Root

Evaluate √81.

Answer

The 𝑛th root, π‘Ÿ, of a quantity 𝑧 is a value such that 𝑧=π‘Ÿ.

It is, therefore, the inverse function to raising a number to the 𝑛th exponent. The 𝑛th root is denoted π‘Ÿ=βˆšπ‘§.

Here, we are asked to find √81; therefore, we need to find the positive value of π‘Ÿ such that 81=π‘Ÿ.οŠͺ

We recall that 3=3Γ—3Γ—3Γ—3=81.οŠͺ

Therefore, π‘Ÿ=3, and we can write √81=3.

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: √27=√.

Answer

We can evaluate the left side of this equation first, √27. An 𝑛th root, π‘Ÿ, of a quantity 𝑧 is a value such that 𝑧=π‘Ÿ.

It is the inverse function to raising a number to the 𝑛th exponent. The 𝑛th root is denoted π‘Ÿ=βˆšπ‘§.

In this case, the value of 𝑛=3 as we are finding the 3rd root. To find the value of π‘Ÿ, where π‘Ÿ=√27, this is equivalent to the value π‘Ÿ such that 27=π‘Ÿ.

As we know that 27=3, we can also say that √27=3.

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: √27=βˆšο‡Œο†²ο‡ο†²ο‡Ž.

For the right-hand side of this equation to be equal to 3, we need to find the missing value such that √=3.

We recall that 3Γ—3=9.

Therefore, √9=3.

Hence, the missing value is 9, and the completed equation is √27=√9.

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 𝑛th Roots

Given that π‘₯=√32βˆ’βˆš625+√64, find π‘₯.

Answer

In order to find the value of π‘₯, we will need to evaluate the 𝑛th roots within this equation.

An 𝑛th root, π‘Ÿ, of a quantity 𝑧 is a value such that 𝑧=π‘Ÿ.

It is the inverse function to raising a number to the 𝑛th exponent. The 𝑛th root is denoted π‘Ÿ=βˆšπ‘§.

Beginning with the first term on the right side of this equation, √32, we can write π‘Ÿ=βˆšπ‘§ο‘ƒ as π‘Ÿ=√32. Then, we find the positive value of π‘Ÿ such that 32=π‘ŸοŠ«. 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 π‘Ÿ. We remember that any integer exponent of 1 will simply be 1; that is, 1=1; so we could use 2 as the first trial for π‘Ÿ.

Since 2Γ—2Γ—2Γ—2Γ—2=32, therefore 2=32.

This means that the 5th root of 32 is 2, so we can write √32=2.

We can apply the same method to find the value of the next term, √625. We will need to find the positive value of π‘Ÿ such that 625=π‘Ÿ.οŠͺ

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 π‘Ÿ must also be an odd number.

We can establish that 3=81,π‘Ÿβ‰ 3.οŠͺso

Next, we can trial the next odd value, π‘Ÿ=5. This gives us 5=625,π‘Ÿ=5.οŠͺso

Therefore, √625=5.

For the final term, √64, we can recall that since 8Γ—8=64, we can write that √64=8.

We can now substitute the values √32=2, √625=5, and √64=8 into the equation to find the value of π‘₯, noting that the term √625 is being subtracted: π‘₯=√32βˆ’βˆš625+√64π‘₯=2βˆ’5+8.

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 π‘₯=5.

We will now consider finding the 𝑛th roots of negative values.

For example, is there a real value π‘₯ for which βˆšβˆ’9=π‘₯?

We know that we can square a positive and negative value in the following example: 3=9(βˆ’3)=9.and

However, both of these give a result of 9, not βˆ’9. So, neither 3 nor βˆ’3 is a solution to βˆšβˆ’9.

Is there a real value π‘Ÿ for which π‘Ÿ=βˆ’9?

If π‘Ÿ is a positive value, then π‘Ÿ=π‘ŸΓ—π‘ŸοŠ¨ would be a positive multiplied by a positive, giving a positive value. Thus π‘Ÿ>0, and not βˆ’9.

If π‘Ÿ is a negative value, then π‘Ÿ=π‘ŸΓ—π‘ŸοŠ¨ would be a negative multiplied by a negative, also giving a positive value, and not βˆ’9.

Thus, there is no real value of π‘₯ for which π‘₯=βˆ’9, and no real solution for βˆšβˆ’9. 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 𝑛 in the 𝑛th root of a negative number.

We consider that 3=3Γ—3Γ—3Γ—3=81,οŠͺ and (βˆ’3)=βˆ’3Γ—βˆ’3ο‡Œο†²ο†²ο‡ο†²ο†²ο‡ŽΓ—βˆ’3Γ—βˆ’3ο‡Œο†²ο†²ο‡ο†²ο†²ο‡Ž=81.οŠͺ

Therefore, if we consider the 4th root of βˆ’81, is it possible to have a real value π‘₯ for which π‘₯=βˆ’81βˆšβˆ’81=π‘₯?οŠͺand

The answer is no, as the only two potential solutions, π‘₯=3 and π‘₯=βˆ’3, both give an answer of positive 81 when raised to the 4th exponent.

Next, we can examine the situations where there are 𝑛th roots of negative numbers.

For example, are there any real values of π‘₯ for which π‘₯=βˆšβˆ’8?

We can calculate that (βˆ’2)=βˆ’2Γ—βˆ’2Γ—βˆ’2=βˆ’8.

Therefore, οŽ’βˆšβˆ’8=βˆ’2.

Taking a different calculation, as we can evaluate (βˆ’3)=βˆ’243, we know that οŽ€βˆšβˆ’243=βˆ’3.

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 𝑛th root of a negative number when 𝑛 is a positive, odd integer.

We can summarize the results in the table below.

For the equation 𝑧=π‘ŸοŠ,

𝑧>0𝑧<0
𝑛 is an even positive integerTwo real roots, Β±βˆšπ‘§ο‘ƒNo real roots
𝑛 is an odd positive integerOne real root, ο‘ƒβˆšπ‘§

We will see how to apply this in our next example.

Example 5: Finding an Unknown Variable Using 𝑛th Roots

Given that π‘₯=√64+√81+βˆšβˆ’27βˆ’βˆš16οŽ₯, find π‘₯.

Answer

To find the value of π‘₯, we will need to evaluate the 𝑛th roots within this equation.

An 𝑛th root, π‘Ÿ, of a quantity 𝑧 is a value such that 𝑧=π‘Ÿ.

It is the inverse function to raising a number to the 𝑛th exponent. The 𝑛th root is denoted π‘Ÿ=βˆšπ‘§.

Taking the terms on the left side of the equation, we can evaluate οŽ₯√64 by considering which positive value of π‘Ÿ would give 64=π‘Ÿ.

As 2=64,π‘Ÿ=2,wemusthave and, as such, οŽ₯√64=2.

Since 81=9Γ—9, we can conclude that √81=9.

We then need to evaluate οŽ’βˆšβˆ’27. Note that when finding 𝑛th roots of a value, when 𝑛 is an even number, we cannot find a real solution to an 𝑛th 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 𝑛 is odd, we can find a real solution to an 𝑛th root of a negative number.

Since we have 3Γ—3Γ—3=3=27, we know that βˆ’3Γ—βˆ’3Γ—βˆ’3=(βˆ’3)=βˆ’27.

Therefore, οŽ’βˆšβˆ’27=βˆ’3.

The final term to evaluate is √16. So, which positive value of π‘Ÿ solves the equation π‘Ÿ=16οŠͺ?

We can write that 2=16.οŠͺ

Therefore, √16=2.

We can now solve the equation π‘₯=√64+√81+βˆšβˆ’27βˆ’βˆš16,οŽ₯ by using the values οŽ₯√64=2, √81=9, οŽ’βˆšβˆ’27=βˆ’3, and √16=2. Substituting these values for the terms, and then simplifying, gives π‘₯=√64+√81+βˆšβˆ’27βˆ’βˆš16π‘₯=2+9+(βˆ’3)βˆ’2π‘₯=6.οŽ₯

The solution for the value of π‘₯ is 6.

Key Points

  • The 𝑛th root of a value is denoted π‘Ÿ=βˆšπ‘§ο‘ƒ. It is the inverse operation to raising a number to the 𝑛th exponent, and we can consider this as finding the value π‘Ÿ, such that 𝑧=π‘ŸοŠ.
  • We can find a real solution to the 𝑛th root of a negative number when 𝑛 is odd.
  • We cannot find a real solution to the 𝑛th root of a negative number when 𝑛 is even.

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