Lesson Explainer: Multiplying and Dividing Square Roots Mathematics • 10th Grade

In this explainer, we will learn how to multiply and divide square roots and use that to simplify expressions.β€Œ

We can begin by recalling that βˆšπ‘Ž is the nonnegative number such that ο€Ίβˆšπ‘Žο†=π‘ŽοŠ¨. In particular, this tells us that if π‘Žβ‰₯0, then ο€Ίβˆšπ‘Žο†=βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž.

This gives us the following result.

Property: Product of the Square Root of a Nonnegative Real Number

If π‘Ž is a nonnegative real number, then βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž.

We can use this property, alongside the commutative and associative properties of multiplication, to simplify expressions involving calculating the product of the square root of a real number with itself. For example, 2√3Γ—6√3=(2Γ—6)Γ—ο€»βˆš3Γ—βˆš3=12Γ—3=36.

Let’s now see another example where we need to use this property to simplify an expression.

Example 1: Multiplying Two Radical Numbers with the Same Base

Calculate 12√3Γ—7√3.

Answer

We can rewrite this expression by using the commutative and associative properties of the multiplication of real numbers as follows: 12√3Γ—7√3=(12Γ—7)Γ—ο€»βˆš3Γ—βˆš3.

We can then recall that, for any nonnegative real number π‘Ž, βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž. Hence, (12Γ—7)Γ—ο€»βˆš3Γ—βˆš3=84Γ—3=252.

Let’s see another example of using this property to simplify an expression.

Example 2: Multiplying a Radical Number and a Radical Expression with the Same Base

Simplify 5√7ο€»1+√7.

Answer

We start by distributing the product over the parentheses to get 5√7ο€»1+√7=ο€»5√7Γ—1+ο€»5√7Γ—βˆš7.

We can simplify this expression by recalling that 1 is the multiplicative identity, so multiplying by 1 leaves the value unchanged, and we can use the associativity of multiplication to rewrite the second term as ο€»5√7Γ—1+ο€»5√7Γ—βˆš7=5√7+5Γ—ο€»βˆš7Γ—βˆš7.

We can then simplify this expression by recalling that, for any nonnegative real number π‘Ž, βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž. Hence, 5√7+5Γ—ο€»βˆš7Γ—βˆš7=5√7+(5Γ—7)=5√7+35.

In our next example, we will simplify the product of two radical expressions.

Example 3: Multiplying two Radical Expressions and Simplifying

Express ο€»5√10+3√35√10βˆ’3√3 in its simplest form.

Answer

We start by distributing over the parentheses to get ο€»5√10+3√35√10βˆ’3√3=ο€»5√10Γ—5√10+ο€»5√10Γ—ο€»βˆ’3√3+ο€»3√3Γ—5√10+ο€»3√3Γ—ο€»βˆ’3√3.

Taking out the factors of βˆ’1 and canceling like terms gives ο€»5√10Γ—5√10+ο€»5√10Γ—ο€»βˆ’3√3+ο€»3√3Γ—5√10+ο€»3√3Γ—ο€»βˆ’3√3=ο€»5√10Γ—5√10ο‡βˆ’ο€»5√10Γ—3√3+ο€»3√3Γ—5√10ο‡βˆ’ο€»3√3Γ—3√3.

We can reorder the second term and simplify to get ο€»5√10Γ—5√10ο‡βˆ’ο€»3√3Γ—5√10+ο€»3√3Γ—5√10ο‡βˆ’ο€»3√3Γ—3√3=ο€»5√10Γ—5√10ο‡βˆ’ο€»3√3Γ—3√3.

We then rearrange the expressions and evaluate to get ο€»5√10Γ—5√10ο‡βˆ’ο€»3√3Γ—3√3=(5Γ—5)Γ—ο€»βˆš10Γ—βˆš10ο‡βˆ’(3Γ—3)Γ—ο€»βˆš3Γ—βˆš3=25ο€»βˆš10Γ—βˆš10ο‡βˆ’9ο€»βˆš3Γ—βˆš3.

We can then simplify this expression by recalling that, for any nonnegative real number π‘Ž, βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž. Hence, 25ο€»βˆš10Γ—βˆš10ο‡βˆ’9ο€»βˆš3Γ—βˆš3=(25Γ—10)βˆ’(9Γ—3)=250βˆ’27=223.

We have demonstrated how to manipulate and evaluate expressions involving products of square roots. We can then ask, β€œWhat happens when we multiply the square roots of any two real numbers?” Of course, we cannot take the square root of negative numbers, so we will begin by assuming that our numbers are nonnegative real numbers, π‘Ž and 𝑏. We can then note that we know how to simplify βˆšπ‘ŽΓ—βˆšπ‘Ž and βˆšπ‘Γ—βˆšπ‘, which we can use to see the following: ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡Γ—ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=ο€Ίβˆšπ‘ŽΓ—βˆšπ‘Žο†Γ—ο€»βˆšπ‘Γ—βˆšπ‘ο‡=π‘ŽΓ—π‘.

We can then take the square root of both sides of the equation. Then, by using the fact that π‘Ž and 𝑏 are nonnegative, we have that βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.

We have therefore shown the following property.

Property: Product of Square Roots of Nonnegative Real Numbers

If π‘Ž and 𝑏 are nonnegative real numbers, then βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.

We can use this result to simplify expressions involving the product of the square roots of nonnegative real numbers. For example, √6Γ—βˆš24=√6Γ—24=√144=12.

We can follow a similar process to evaluate the quotient of two square roots. This time we assume that π‘Ž and 𝑏 are nonnegative and that 𝑏 is nonzero. We can then apply the same reasoning as before to see the following: ο€Ώβˆšπ‘Žβˆšπ‘ο‹=βˆšπ‘Žβˆšπ‘Γ—βˆšπ‘Žβˆšπ‘=βˆšπ‘ŽΓ—βˆšπ‘Žβˆšπ‘Γ—βˆšπ‘=π‘Žπ‘.

Taking square roots of both sides of the equation gives us the following result.

Property: Square Root of the Quotient of Nonnegative Real Numbers

If π‘Ž and 𝑏 are nonnegative real numbers and 𝑏≠0, then βˆšπ‘Žβˆšπ‘=ο„žπ‘Žπ‘.

We can use this process to simplify expressions involving the quotient of the square roots of nonnegative real numbers. For example, √18√2=ο„ž182=√9=3.

It is worth noting that both of these results also work in the other direction. For example, ο„ž254=√25√4=52.

Let’s now see an example of how to use these properties to write an expression involving radicals in its simplest form.

Example 4: Squaring a Radical Expression and Simplifying

Express ο€»βˆš7+√3ο‡βˆ’βˆš84 in its simplest form.

Answer

We first distribute the exponent over the parentheses to get ο€»βˆš7+√3ο‡βˆ’βˆš84=ο€»βˆš7+√3ο‡ο€»βˆš7+√3ο‡βˆ’βˆš84=ο€»βˆš7Γ—βˆš7+ο€»βˆš7Γ—βˆš3+ο€»βˆš3Γ—βˆš7+ο€»βˆš3Γ—βˆš3ο‡βˆ’βˆš84.

We can reorder the third term and simplify as follows: ο€»βˆš7Γ—βˆš7+ο€»βˆš7Γ—βˆš3+ο€»βˆš7Γ—βˆš3+ο€»βˆš3Γ—βˆš3ο‡βˆ’βˆš84=ο€»βˆš7Γ—βˆš7+2ο€»βˆš7Γ—βˆš3+ο€»βˆš3Γ—βˆš3ο‡βˆ’βˆš84.

We then recall that if π‘Ž is a nonnegative real number, then βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž. This means that we can simplify this expression as follows: ο€»βˆš7Γ—βˆš7+2ο€»βˆš7Γ—βˆš3+ο€»βˆš3Γ—βˆš3ο‡βˆ’βˆš84=7+2ο€»βˆš7Γ—βˆš3+3βˆ’βˆš84=10+2ο€»βˆš7Γ—βˆš3ο‡βˆ’βˆš84.

We can simplify further by recalling that if π‘Ž and 𝑏 are nonnegative real numbers, then βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘. We can apply this in two different ways to this expression.

First, √7Γ—βˆš3=√7Γ—3=√21.

Second, we can factor the radicand 84 into primes to get 84=2Γ—2Γ—3Γ—7=4Γ—21. We can use this to simplify the radical: √84=√4Γ—21=√4Γ—βˆš21=2√21.

Hence, 10+2ο€»βˆš7Γ—βˆš3ο‡βˆ’βˆš84=10+2√21βˆ’2√21=10.

In our next example, we will simplify an expression involving the quotient of two radicals with different bases.

Example 5: Dividing Two Radical Numbers with Different Bases

Express √10√2 in its simplest form.

Answer

We first recall that if π‘Ž and 𝑏 are nonnegative real numbers and 𝑏≠0, then βˆšπ‘Žβˆšπ‘=ο„žπ‘Žπ‘. Applying this to the given expression yields √10√2=ο„ž102.

We note that 10 and 2 share a common factor of 2. Canceling this shared factor yields ο„ž102=ο„ž5Γ—21Γ—2=√5.

In our next example, we will simplify an expression using all of the properties we have found in this explainer.

Example 6: Simplifying an Expression Involving the Multiplication and Division of Radicals

Express √2√8Γ—ο€»βˆš20+√36 in its simplest form.

Answer

We start by distributing over the parentheses to get √2√8Γ—ο€»βˆš20+√36=ο€Ώβˆš2√8Γ—βˆš20+ο€Ώβˆš2√8Γ—βˆš36=√2Γ—βˆš20√8+√2Γ—βˆš36√8.

We then recall that if π‘Ž is a nonnegative real number, then βˆšπ‘Ž=π‘ŽοŠ¨, and if π‘Ž and 𝑏 are nonnegative real numbers, then βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘. Applying this to the expression gives √2Γ—βˆš20√8+√2Γ—βˆš36√8=√2Γ—20√8+√2Γ—βˆš6√8=√2Γ—10√8+6√2√8=2√10√8+6√2√8.

We then recall that if π‘Ž and 𝑏 are nonnegative real numbers and 𝑏≠0, then βˆšπ‘Žβˆšπ‘=ο„žπ‘Žπ‘. Applying this we get 2√10√8+6√2√8=2ο„ž108+6ο„ž28=2ο„ž54+6ο„ž14.

We then note that ο„ž54=√5√4=√52 and ο„ž14=√1√4=12, so 2ο„ž54+6ο„ž14=2√52+62=√5+3=3+√5.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • If π‘Ž is a nonnegative real number, then βˆšπ‘ŽΓ—βˆšπ‘Ž=π‘Ž.
  • If π‘Ž and 𝑏 are nonnegative real numbers, then βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘.
  • If π‘Ž and 𝑏 are nonnegative real numbers and 𝑏≠0, then βˆšπ‘Žβˆšπ‘=ο„žπ‘Žπ‘.

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