Lesson Explainer: nth Roots: Expressions and Equations Mathematics

In this explainer, we will learn how to simplify algebraic expressions and solve algebraic equations involving ๐‘›th roots, where ๐‘› is a positive integer greater than or equal to 2.

The ๐‘›th root is an important mathematical operation that describes the inverse of a power operation with an exponent ๐‘›. Letโ€™s begin by providing a formal definition for the ๐‘›th root.

Definition: ๐‘›th Roots

The ๐‘›th root of a number ๐‘ฅ, where ๐‘› is a positive integer, is a number that, when raised to the ๐‘›th power, gives ๐‘ฅ. We can define this number as ๐‘ฆ, such that ๐‘ฅ=๐‘ฆ.๏Š

The ๐‘›th root is given by ๐‘ฆ=โˆš๐‘ฅ๏‘ƒ.

Note that the ๐‘›th root of ๐‘ฅ can equivalently be written as ๐‘ฅ๏Ž ๏‘ƒ. It is outside the scope of this lesson to express ๐‘›th roots using this notation, but if you already are familiar with the notation, it can act as a helpful tool to understanding how the rules of exponents can be applied to expressions involving roots.

Theorem: Properties of ๐‘›th Roots

  • When ๏‘ƒโˆš๐‘Ž and ๏‘ƒโˆš๐‘ are well defined and are real numbers, then ๏‘ƒโˆš๐‘Ž๐‘ is also defined such that ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Žโˆš๐‘=โˆš๐‘Ž๐‘.
  • If ๏‘ƒโˆš๐‘โ‰ 0, then it is also the case that ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Žโˆš๐‘=๏„ž๐‘Ž๐‘.
  • When ๐‘› is an odd integer, ๏€บโˆš๐‘Ž๏†=โˆš๐‘Ž=๐‘Ž.๏‘ƒ๏‘ƒ๏Š๏Š
  • When ๐‘› is an even integer and ๐‘Žโ‰ฅ0, ๏€บโˆš๐‘Ž๏†=๐‘Ž.๏‘ƒ๏Š
  • When ๐‘› is an even integer and ๐‘Ž<0, ๏€บโˆš๐‘Ž๏†โ„.๏‘ƒ๏Šisunde๏ฌnedover
  • When ๐‘› is an even integer and ๐‘Žโˆˆโ„, ๏‘ƒโˆš๐‘Ž=|๐‘Ž|.๏Š

In our first two examples, we will demonstrate how to apply a combination of these properties to simplify an expression involving a root.

Example 1: Simplifying Algebraic Expressions Involving Exponents and Cube Roots

Simplify ๏Žขโˆš64๐‘š๏Šฉ.

Answer

Recall that, for positive real numbers ๐‘Ž and ๐‘ and positive integers ๐‘›, ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Žโˆš๐‘=โˆš๐‘Ž๐‘.

By applying this property in reverse, we can rewrite ๏Žขโˆš64๐‘š๏Šฉ as ๏Žข๏Žข๏Žขโˆš64๐‘š=โˆš64โˆš๐‘š.๏Šฉ๏Šฉ

Next, we know that, for odd integers ๐‘›, ๏€บโˆš๐‘Ž๏†=โˆš๐‘Ž=๐‘Ž.๏‘ƒ๏‘ƒ๏Š๏Š

Hence, ๏Žขโˆš๐‘š=๐‘š.๏Šฉ

Similarly, since 64=4๏Šฉ, ๏Žข๏Žขโˆš64=โˆš4=4.๏Šฉ

Combining these expressions, we obtain ๏Žขโˆš64๐‘š=4๐‘š.๏Šฉ

Example 2: Simplifying Algebraic Expressions Involving Exponents and Square Roots

Simplify โˆš100๐‘ฅ๏Šง๏Šฌ.

Answer

When a root of the form ๏‘ƒโˆš๐‘ฅ is given with ๐‘› omitted, we assume ๐‘›=2. This means we can simplify โˆš100๐‘ฅ๏Šง๏Šฌ by applying the property of ๐‘›th roots with ๐‘›=2, โˆš๐‘Žโˆš๐‘=โˆš๐‘Ž๐‘.

Thus, โˆš100๐‘ฅ=โˆš100โˆš๐‘ฅ=10โˆš๐‘ฅ.๏Šง๏Šฌ๏Šง๏Šฌ๏Šง๏Šฌ

Finally, by writing ๐‘ฅ๏Šง๏Šฌ as ๏€น๐‘ฅ๏…๏Šฎ๏Šจ and using the property ๏‘ƒโˆš๐‘ฅ=๐‘ฅ๏Š for even ๐‘›, we can simplify โˆš๐‘ฅ๏Šง๏Šฌ as follows: โˆš๐‘ฅ=๏„(๐‘ฅ)=๐‘ฅ.๏Šง๏Šฌ๏Šฎ๏Šจ๏Šฎ

Hence, โˆš100๐‘ฅ=10๐‘ฅ.๏Šง๏Šฌ๏Šฎ

Great care must be taken when finding powers of roots, such as ๏‘ƒโˆš๐‘ฅ๏‰ or ๏€บโˆš๐‘ฅ๏†๏‘ƒ๏‰. In the previous example, we demonstrated that ๏„(๐‘ฅ)=๐‘ฅ๏Šฎ๏Šจ๏Šฎ. Since we had an even power inside an even root, the operation was defined for all real values of ๐‘ฅ. If, however, we were simplifying โˆš๐‘ฅ๏Šง๏Šช, we would need to perform the following steps: โˆš๐‘ฅ=๏„(๐‘ฅ)=||๐‘ฅ||.๏Šง๏Šช๏Šญ๏Šจ๏Šญ

In this case, ๐‘ฅ๏Šง๏Šช inside the square root acts first, ensuring the radicand is positive for all values of ๐‘ฅ. This means we are not taking the even root of a negative number and, in turn, means the resulting function can only output positive values. Since ๐‘ฅ๏Šญ is negative for values of ๐‘ฅ<0, we must include the absolute value when simplifying, as shown.

In the last two examples, we used a property of roots to express an ๐‘›th root as the product of two unique ๐‘›th roots. It is important to realize that we can extend this property to express a root as the product of three or more ๐‘›th roots in order to simplify an expression. In other words, for positive real numbers ๐‘Ž, ๐‘, and ๐‘ and positive integers ๐‘›, ๏‘ƒ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Ž๐‘๐‘=โˆš๐‘Žโˆš๐‘โˆš๐‘.

Example 3: Simplifying Algebraic Expressions with More Than One Variable Involving Exponents and Square Roots

Write โˆš25๐‘Ž๐‘๏Šจ๏Šฌ in its simplest form.

Answer

Recall that, for positive real numbers ๐‘Ž, ๐‘, and ๐‘ and positive integers ๐‘›, ๏‘ƒ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Ž๐‘๐‘=โˆš๐‘Žโˆš๐‘โˆš๐‘.

Since โˆš25๐‘Ž๐‘๏Šจ๏Šฌ is equivalent to ๏Žกโˆš25๐‘Ž๐‘๏Šจ๏Šฌ, we can rewrite it as shown: โˆš25๐‘Ž๐‘=โˆš25โˆš๐‘Žโˆš๐‘=5โˆš๐‘Žโˆš๐‘.๏Šจ๏Šฌ๏Šจ๏Šฌ๏Šจ๏Šฌ

Next, by using the property ๏‘ƒโˆš๐‘ฅ=|๐‘ฅ|๏Š for even ๐‘›, we can write ๏Žกโˆš๐‘Ž=|๐‘Ž|.๏Šจ

Similarly, ๐‘๏Šฌ can be written as ๏€น๐‘๏…๏Šฉ๏Šจ, so โˆš๐‘=๏„(๐‘)=||๐‘||.๏Šฌ๏Šฉ๏Šจ๏Šฉ

This means โˆš25๐‘Ž๐‘=5|๐‘Ž|||๐‘||๏Šจ๏Šฌ๏Šฉ. Since 5>0, 5=|5|, and the product of absolute value is the absolute value of the products, โˆš25๐‘Ž๐‘=||5๐‘Ž๐‘||.๏Šจ๏Šฌ๏Šฉ

In the previous examples, we demonstrated how to use properties of roots to simplify expressions. Letโ€™s now consider an equation involving exponents, ๐‘ฆ=16.๏Šจ

A solution to this equation is found by taking the square root, such that ๐‘ฆ=โˆš16=4.

However, if we substitute ๐‘ฆ=โˆ’4 into the expression ๐‘ฆ๏Šจ, we obtain (โˆ’4)=16๏Šจ. This means that ๐‘ฆ=โˆ’4 is also a solution to the equation ๐‘ฆ=16๏Šจ. Therefore, when solving an equation ๐‘ฆ=๐‘ฅ๏Šจ, the solutions include both the positive and negative square roots of ๐‘ฅ. This gives rise to a subtle difference between the two seemingly equivalent statements for integer ๐‘›, ๐‘ฆ=๐‘ฅ๐‘ฆ=โˆš๐‘ฅ.๏Šand๏‘ƒ

We can generalize this concept for ๐‘›th roots where ๐‘› is an even number.

Theorem: Even and Odd ๐‘›th Roots

Consider the equation ๐‘ฆ=๐‘ฅ๏Š for real numbers ๐‘ฆ and ๐‘ฅ and positive integers ๐‘›. The following solutions hold:

๐‘› Even๐‘› Odd
๐‘ฅ<0There are no real solutions to the equation.There is one solution,
๐‘ฆ=โˆš๐‘ฅ๏‘ƒ.
๐‘ฅ>0The solutions to the equation are
๐‘ฆ=ยฑโˆš๐‘ฅ๏‘ƒ.

We have seen that we interpret the statements ๐‘ฆ=๐‘ฅ๏Š and ๐‘ฆ=โˆš๐‘ฅ๏‘ƒ differently. This gives rise to an additional definition, that of the principal ๐‘›th root. This allows us to consider an ๐‘›th root as a function by making it, by definition, a one-to-one mapping.

Definition: Principal ๐‘›th Root

Every positive real number has a single positive ๐‘›th root, defined ๏‘ƒโˆš๐‘ฅ. This is known as the principal ๐‘›th root.

Letโ€™s demonstrate an application of the properties of even and odd ๐‘›th roots in the next example.

Example 4: Simplifying and Solving Equations Involving ๐‘›th Roots

Find the value (or values) of ๐‘ฅ if 12๐‘ฅ=384๏Šซ.

Answer

To solve this equation, we will apply a series of inverse operations. First, we divide both sides by 12: 12๐‘ฅ12=38412๐‘ฅ=32.๏Šซ๏Šซ

Next, we recall that, given an equation ๐‘ฅ=๐‘ฆ๏Š for real numbers ๐‘ฅ and ๐‘ฆ and positive integers ๐‘›,

  • if ๐‘› is even, the solutions to the equation are ๐‘ฅ=ยฑโˆš๐‘ฆ๏‘ƒ,
  • if ๐‘› is odd, there is just one solution, ๐‘ฅ=โˆš๐‘ฆ๏‘ƒ.

Since ๐‘›=5 is odd, there is one solution to this equation, given by ๐‘ฅ=โˆš32=2๏Žค.

Hence, ๐‘ฅ=2.

By combining the properties of ๐‘›th roots and this theorem, we are able solve more complicated equations involving exponents. Letโ€™s demonstrate this with one final example.

Example 5: Simplifying and Solving Equations Involving ๐‘›th Roots

Find the value (or values) of ๐‘ฅ given that ๏€ผ๐‘ฅ+95๏ˆ=โˆš144ร—3๏Šจ๏Šจ.

Answer

We will begin by evaluating the right-hand side of this equation. Since 144 and 3๏Šจ are both square numbers, we can easily find the principal square root of their product, which is โˆš144ร—3=โˆš144โˆš3=12ร—3=36.๏Šจ๏Šจ

Next, we recall that, given an equation ๐‘ฅ=๐‘Ž๏Š for real numbers ๐‘ฅ and ๐‘Ž and positive integers ๐‘›, if ๐‘› is even, the solutions to the equation are ๐‘ฅ=ยฑโˆš๐‘Ž๏‘ƒ. Since the expression on the left-hand side of the equation has an even exponent (๐‘›=2), we will need to take the positive and negative square root of 36.

In particular, ๏€ผ๐‘ฅ+95๏ˆ=โˆš144ร—3๏€ผ๐‘ฅ+95๏ˆ=36๐‘ฅ+95=ยฑโˆš36๐‘ฅ+95=ยฑ6.๏Šจ๏Šจ๏Šจ

To find the solutions to these two equations, we will next multiply both sides by 5 and then subtract 9: ๐‘ฅ+95=6๐‘ฅ+95=โˆ’6๐‘ฅ+9=30๐‘ฅ+9=โˆ’30๐‘ฅ=21,๐‘ฅ=โˆ’39.

Hence, the solutions are ๐‘ฅ=21 and ๐‘ฅ=โˆ’39.

Letโ€™s finish by recapping some key concepts from this explainer.

Key Points

  • The ๐‘›th root (or radical) of a number ๐‘ฅ, where ๐‘› is a positive integer, is a number that, when raised to the ๐‘›th power, gives ๐‘ฅ. We can define this number as ๐‘ฆ, such that ๐‘ฅ=๐‘ฆ.๏Š
  • The ๐‘›th root is given by ๐‘ฆ=โˆš๐‘ฅ๏‘ƒ.
  • Every positive real number has a single positive ๐‘›th root, defined ๏‘ƒโˆš๐‘ฅ. This is known as the principal ๐‘›th root.
  • When ๏‘ƒโˆš๐‘Ž and ๏‘ƒโˆš๐‘ are well defined and are real numbers, ๏‘ƒโˆš๐‘Ž๐‘ is also defined such that ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Žโˆš๐‘=โˆš๐‘Ž๐‘.
  • If ๏‘ƒโˆš๐‘โ‰ 0, then it is also the case that ๏‘ƒ๏‘ƒ๏‘ƒโˆš๐‘Žโˆš๐‘=๏„ž๐‘Ž๐‘.
  • When ๐‘› is an odd integer, ๏€บโˆš๐‘Ž๏†=โˆš๐‘Ž=๐‘Ž.๏‘ƒ๏‘ƒ๏Š๏Š
  • When ๐‘› is an even integer and ๐‘Ž>0, ๏€บโˆš๐‘Ž๏†=๐‘Ž.๏‘ƒ๏Š
  • When ๐‘› is an even integer and ๐‘Ž<0, ๏€บโˆš๐‘Ž๏†โ„.๏‘ƒ๏Šisunde๏ฌnedover
  • When ๐‘› is an even integer and ๐‘Žโˆˆโ„, ๏‘ƒโˆš๐‘Ž=|๐‘Ž|.๏Š
  • The solutions to the equation ๐‘ฆ=๐‘ฅ๏Š for real numbers ๐‘ฅ and ๐‘ฆ and positive integers ๐‘› are as follows:
    ๐‘› Even๐‘› Odd
    ๐‘ฅ<0There are no real solutions to the equation.There is one solution,
    ๐‘ฆ=โˆš๐‘ฅ๏‘ƒ.
    ๐‘ฅ>0The solutions to the equation are ๐‘ฆ=ยฑโˆš๐‘ฅ๏‘ƒ.

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