Lesson Explainer: Adding and Subtracting Square Roots | Nagwa Lesson Explainer: Adding and Subtracting Square Roots | Nagwa

Lesson Explainer: Adding and Subtracting Square Roots Mathematics • Second Year of Preparatory School

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In this explainer, we will learn how to add and subtract square roots and how to use that to simplify expressions.

Before we can simplify radical expressions (i.e., expressions containing roots), we first need to consider what it means to simplify a radical expression and which expressions can actually be simplified. For example, we can note that, for any real number π‘₯, the additive inverse property tells us that the number βˆ’π‘₯ is real and that π‘₯+(βˆ’π‘₯)=0, or more succinctly, π‘₯βˆ’π‘₯=0. We can use this to simplify radical expressions: √2βˆ’βˆš2=0.

We can combine this with the distributive property of multiplication over addition to combine radical expressions. For example, we recall that, for any real numbers π‘Ž, 𝑏, and 𝑐, we have π‘Žπ‘+𝑐𝑏=(π‘Ž+𝑐)𝑏. Therefore, if we want to evaluate 3√2+2√2, we can take out the shared factor of √2 to get 3√2+2√2=(3+2)√2=5√2.

Let’s see an example of applying this property to simplify an expression involving radicals.

Example 1: Simplifying the Addition and Subtraction of Square Roots

Simplify ο€»βˆ’2+√6+ο€»8βˆ’βˆš6.

Answer

We first recall that the addition of real numbers is associative and commutative, so we can add the terms in this expression in any order. Thus, ο€»βˆ’2+√6+ο€»8βˆ’βˆš6=(βˆ’2+8)+ο€»βˆš6βˆ’βˆš6=6+ο€»βˆš6βˆ’βˆš6.

We then recall that the additive identity property of the addition of real numbers tells us that, for any real number π‘₯, we have π‘₯βˆ’π‘₯=0. By setting π‘₯=√6, we can see that √6βˆ’βˆš6=0. Hence, 6+ο€»βˆš6βˆ’βˆš6=6+0=6.

This method for the simplification of square roots allows us to simplify the roots of nonnegative integers and to use the additive inverse property of addition to simplify the addition and subtraction of radicals with the same base. There is another way of simplifying radical expressions, and we will see this with the following example.

Consider 2√3+√3. We know that 3 has no square divisors greater that 1, so we cannot simplify either of the terms. Instead, we can recall that, for any real number π‘₯, we have 2π‘₯=π‘₯+π‘₯.

In particular, if π‘₯=√3, we have 2√3=√3+√3.

We can therefore rewrite the expression as follows: 2√3+√3=ο€»βˆš3+√3+√3.

We can then note that, for any real number π‘₯, we have 3π‘₯=π‘₯+π‘₯+π‘₯. Hence, ο€»βˆš3+√3+√3=3√3.

This gives us a link between adding and subtracting square roots with the same base and integer multiplication. We have the following result.

Property: Addition and Subtraction of Integer Multiples of Square Roots with the Same Base

For any integers π‘Ž, 𝑏, and 𝑐, where 𝑐 is nonnegative, we have π‘Žβˆšπ‘+π‘βˆšπ‘=(π‘Ž+𝑏)βˆšπ‘.

This result also holds true for any real numbers π‘Ž and 𝑏 by considering the distributive property of the multiplication of real numbers over the addition of real numbers, which states that, for any real numbers π‘Ž, 𝑏, and 𝑐, we have π‘ŽΓ—(𝑏+𝑐)=(π‘ŽΓ—π‘)+(π‘ŽΓ—π‘).

Let’s now see an example of applying this property to simplify the sum of radicals with the same base.

Example 2: Simplifying the Addition of Two Square Roots with the Same Base

Simplify 12√5+3√5.

Answer

We recall that, for any integers π‘Ž, 𝑏, and 𝑐, where 𝑐 is nonnegative, we have π‘Žβˆšπ‘+π‘βˆšπ‘=(π‘Ž+𝑏)βˆšπ‘.

Therefore, 12√5+3√5=(12+3)√5=15√5.

Hence, 12√5+3√5=15√5.

A key point to note is that we cannot just add or subtract the bases of the square roots. Instead, we are simplifying the coefficients.

Consider 5+6√3. A very common error is to simplify this to 11√3, whereas, in reality, the expression cannot be simplified further. Similarly, a common error is to simplify √2+√3 as √5. Once again, we cannot combine these radicals since the radicands (the expressions under the root) are different.

In our next example, we will consider simplifying the sum of two radical expressions.

Example 3: Simplifying the Addition of Two Algebraic Expressions Involving Square Roots

Given that π‘Ž=6βˆ’7√3 and 𝑏=βˆ’7βˆ’7√3, find the value of π‘Ž+𝑏.

Answer

We first substitute the expressions for π‘Ž and 𝑏 to see that π‘Ž+𝑏=ο€»6βˆ’7√3+ο€»βˆ’7βˆ’7√3.

We can then use the commutativity and associativity of the addition of real numbers to reorder the addition to the following: ο€»6βˆ’7√3+ο€»βˆ’7βˆ’7√3=(6+(βˆ’7))+ο€»βˆ’7√3βˆ’7√3.

We note that 6+(βˆ’7)=6βˆ’7=βˆ’1, and we recall that, for any integers π‘Ž, 𝑏, and 𝑐, where 𝑐 is nonnegative, we have π‘Žβˆšπ‘+π‘βˆšπ‘=(π‘Ž+𝑏)βˆšπ‘. This means that βˆ’7√3βˆ’7√3=(βˆ’7βˆ’7)√3=βˆ’14√3.

Hence, π‘Ž+𝑏=βˆ’1βˆ’14√3.

A further property of square roots is illustrated by the following. We first note that βˆšπ‘ŽΓ—π‘ is the nonnegative number, which when squared gives π‘ŽΓ—π‘. We note that if π‘Ž,𝑏β‰₯0, then ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=βˆšπ‘ŽΓ—βˆšπ‘Γ—βˆšπ‘ŽΓ—βˆšπ‘=ο€Ίβˆšπ‘Žο†Γ—ο€»βˆšπ‘ο‡=π‘ŽΓ—π‘.

Thus, the square of βˆšπ‘ŽΓ—βˆšπ‘ is also π‘ŽΓ—π‘. Since βˆšπ‘ŽΓ—βˆšπ‘ is also nonnegative, we have the following result.

Property: The Product Property of Square Roots

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

Let’s use this property to rewrite √8 by noting that √8=√2Γ—2=√2Γ—βˆš2=2√2.

It is not clear which (if any) of √8 or 2√2 is the simpler expression. We therefore need to define what we mean by a simplified expression involving square roots.

Definition: Simplified Radical Expression

If 𝑐 is a nonnegative integer, we can rewrite βˆšπ‘=π‘Žβˆšπ‘, where π‘Ž is the square root of the largest the perfect square factor of 𝑐 and 𝑏 is the other non-perfect square factor of 𝑐. This is called the simplified form of βˆšπ‘.

It then follows that the simplified form of an expression involving square roots involves writing every term in simplified form. This allows us to say that 2√2 is the simplified form of √8. In fact, this gives us a method of determining the simplified form of the square root of a nonnegative integer.

We can simplify some radical expressions by noting that if π‘Ž>1 and π‘ŽοŠ¨ divides 𝑐, say 𝑐=π‘ŽΓ—π‘οŠ¨, then βˆšπ‘=βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘=π‘Žβˆšπ‘.

This means we can simplify βˆšπ‘ if it has any perfect square divisors greater than 1. This means we determine the largest the perfect square factor of 𝑐 to simplify βˆšπ‘.

In our next example, we will simplify the sum of two radicals where the base (or radicand) of each term is different.

Example 4: Simplifying the Addition of Two Square Roots with Different Radicands

Write √8+√2 in the form π‘Žβˆš2, where π‘Ž is an integer.

Answer

We first note that the base of each square root is different, so we cannot simplify this expression in its current form by addition. Instead, we note that 8=2Γ—2, so we can simplify √8 by using the fact that, for any nonnegative integers π‘Ž and 𝑏, we have βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘. Therefore, √8=√2Γ—βˆš2=2√2.

We can use this to rewrite the expression given in the question as √8+√2=2√2+√2.

We then recall that, for any integers π‘Ž, 𝑏, and 𝑐, where 𝑐 is nonnegative, we have π‘Žβˆšπ‘+π‘βˆšπ‘=(π‘Ž+𝑏)βˆšπ‘.

Hence, 2√2+√2=(2+1)√2=3√2.

Finally, we note that 2 has no perfect square divisors greater than 1, so this cannot be simplified any further.

In our next example, we will simplify the sum of three radical expressions, each with a different radicand.

Example 5: Simplifying the Addition and Subtraction of Three Square Roots with Different Radicands

Simplify √12βˆ’2√3+4√27.

Answer

We first note that each term has a different radicand, so we cannot directly combine the terms of this expression in its current form. Instead, let’s simplify each term separately by using the fact that, for any nonnegative integers π‘Ž and 𝑏, we have βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘.

We first note that 12=2Γ—3, so √12=√2Γ—βˆš3=2√3.

We then note that 3 has no perfect square divisors greater than 1, so √3 cannot be simplified further. This means we cannot simplify this term or the second term any further.

We then find that 27=3Γ—3, so 4√27=4ο€»βˆš3Γ—βˆš3=4ο€»3√3=12√3.

Substituting these values into the given expression yields √12βˆ’2√3+4√27=2√3βˆ’2√3+12√3.

We can then simplify this by noting that βˆ’2√3 is the additive inverse of 2√3, so that 2√3βˆ’2√3=0.

Hence, 2√3βˆ’2√3+12√3=12√3.

Therefore, √12βˆ’2√3+4√27=12√3.

In our final example, we will use this method for the simplification of radical expressions to determine the value of an unknown in an equation.

Example 6: Finding the Missing Value in the Sum of Two Square Roots with Different Radicands

Given that 9√8+10√50=π‘₯√2, find the value of π‘₯.

Answer

Since the right-hand side of the equation is of the form π‘₯√2 for some real value of π‘₯, we will start by trying to write the left-hand side of the equation in this form. We can do this by simplifying each term using the fact that, for any nonnegative integers π‘Ž and 𝑏, we have βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘. Applying this, we see that 9√8=9ο€»βˆš4Γ—βˆš2=9ο€»2√2=18√2,10√50=10ο€»βˆš25Γ—βˆš2=10ο€»5√2=50√2.

Substituting these values into the equation gives 18√2+50√2=π‘₯√2.

We can simplify the left-hand side by factoring out √2 to get (18+50)√2=π‘₯√268√2=π‘₯√2.

Hence, π‘₯=68.

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

Key Points

  • For any integers π‘Ž, 𝑏, and 𝑐, where 𝑐 is nonnegative, we have π‘Žβˆšπ‘+π‘βˆšπ‘=(π‘Ž+𝑏)βˆšπ‘.
  • If π‘Ž and 𝑏 are nonnegative real numbers, then βˆšπ‘ŽΓ—π‘=βˆšπ‘ŽΓ—βˆšπ‘.
  • If 𝑐 is a nonnegative integer, we can rewrite βˆšπ‘=π‘Žβˆšπ‘, where π‘Ž is the square root of the largest the perfect square factor of 𝑐 and 𝑏 is the other non-perfect square factor of 𝑐. This is called the simplified form of βˆšπ‘.
  • If a nonnegative integer 𝑐 has no perfect square divisors greater than 1, then βˆšπ‘ cannot be simplified.

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