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Lesson Explainer: Surds Mathematics

In this explainer, we will learn how to use the properties of surds to simplify expressions.

The square root of a number βˆšπ‘Ž is the nonnegative number that when multiplied by itself returns the original number. This is called the principal square root of π‘Ž. For example, we can calculate that √9=3 since 3Γ—3=9. In fact, the square root of any perfect square is an integer; βˆšπ‘Ž=π‘ŽοŠ¨ for any nonnegative integer π‘Ž.

If π‘Ž is a positive integer that is not a perfect square, then we say that βˆšπ‘Ž is a surd. Surds are irrational numbers; they cannot be written as the quotients of integers. In particular, this tells us that the decimal expansion of a surd is infinite and nonrepeating. We sometimes refer to expressions involving roots as surds (or radicals) and the expression inside the root as the radicand.

Surds appear throughout mathematics, often as exact solutions to equations. For example, if we sketch a square with sides of length 1, then we can use the Pythagorean theorem to show that the diagonals of the square have length √2.

We recall that the Pythagorean theorem tells us that the square of the length of the hypotenuse, in a right triangle, is equal to the sum of the squares of the lengths of the two shorter sides. So, the length of the hypotenuse β„Ž must satisfy β„Ž=1+1=2. Since β„Ž is positive, we have that β„Ž=√2.

We can use the properties and operations of real numbers to evaluate and simplify expressions involving surds. For example, we can subtract surds with the same radicand (number inside the square root): √5βˆ’βˆš5=0.

We can also take out shared factors of surds to simplify expressions: 2√5+3√5βˆ’βˆš5=(2+3βˆ’1)√5=4√5.

There are two key properties that we can use to simplify surds.

Properties: Surds

For any positive numbers π‘Ž and 𝑏, we have that

  1. βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘,
  2. ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘.

We can prove these results by squaring each expression. For example, we can note that since π‘Ž and 𝑏 are positive, ο€»βˆšπ‘Žπ‘ο‡=π‘Žπ‘οŠ¨ and ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡Γ—ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡.

Then, using the commutativity and associativity of multiplication, we can rewrite this as ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡Γ—ο€»βˆšπ‘ŽΓ—βˆšπ‘ο‡=ο€Ίβˆšπ‘ŽΓ—βˆšπ‘Žο†Γ—ο€»βˆšπ‘Γ—βˆšπ‘ο‡=ο€Ίβˆšπ‘Žο†Γ—ο€»βˆšπ‘ο‡=π‘Žπ‘.

Therefore, both βˆšπ‘Žπ‘ and βˆšπ‘ŽΓ—βˆšπ‘ are square roots of π‘Žπ‘ and since they are both positive, they must be equal. A similar argument can be made to prove that ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘ for positive values of π‘Ž and 𝑏.

We can use these properties to simplify surds whose radicands contain perfect square factors. For example, consider √12. We can lower the radicand of this surd by representing 12 as a product of its prime factors; namely, 12=2Γ—3. Therefore, √12=√2Γ—3.

If we substitute π‘Ž=2 and 𝑏=3 into the rule βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘, we obtain √2Γ—3=√2Γ—βˆš3.

We can calculate that √2=2, so √2Γ—βˆš3=2√3.

We cannot reduce the radicand any further since 3 contains no perfect square factors greater than 1. In general, we simplify a surd by writing it in the form π‘βˆšπ‘ž, where π‘ž has no perfect square factors above 1.

In practice, when simplifying surds, we try to identify if the radicand (the number inside the square root) has a factor that is a perfect square, and if it does, then we try to find the largest of these (for the sake of efficiency).

For example, with the example above, rather than writing the prime factorization of 12, we could have used that 12=4Γ—3.

Let’s see an example of using these properties to simplify a given surd.

Example 1: Simplifying a Surd

Simplify √108.

Answer

To simplify a surd, we need to write it in the form π‘βˆšπ‘ž, where π‘ž has no perfect square factors greater than 1. We can do this by recalling that, for any positive numbers π‘Ž and 𝑏, βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘.

If we look for perfect square factors of 108, we can note that 108=4Γ—27 and 108=9Γ—12, and we may even notice that 108=36Γ—3. We can simplify the square root using any of these factorizations. But since 36 is the largest perfect square factor of 108, we will start with this value, as it will simplify our calculation: √108=√36Γ—3=√36Γ—βˆš3.

We note that 36=6, so we have √36Γ—βˆš3=√6Γ—βˆš3=6√3.

Hence, √108=6√3.

In the previous example, we identified the largest perfect square factor of the radicand to simplify the surd. However, it can be difficult to determine all of the perfect square factors. In cases like this, we can always factor the radicand into primes.

If we were to apply this method to the previous question, we would first note that 108 is even and 108=2Γ—54. Next, we would note that 54 is even and 2Γ—54=2Γ—2Γ—27. Now, we can use the fact that 27=3 to see that 108=2Γ—3.

Thus, √108=√2Γ—3.

We have π‘Ž=2 and 𝑏=3, so √2Γ—3=√2Γ—βˆš3.

We can evaluate √2=2 and we can also note that 3=3Γ—3, so we can rewrite the surd as follows: √2Γ—βˆš3=2Γ—βˆš3Γ—3=2Γ—βˆš3Γ—βˆš3.

Finally, we have that √3=3, so 2Γ—βˆš3Γ—βˆš3=2Γ—3Γ—βˆš3=6√3.

It is worth noting that we could have simplified this surd by only applying this once using the fact that 108=36Γ—3=6Γ—3, so √108=√6Γ—3=√6Γ—βˆš3=6√3.

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

Example 2: Simplifying a Product of Surds

Simplify √10Γ—βˆš15.

Answer

To simplify the product of two surds, we could start by simplifying each surd separately; however, in this case, neither radicand has a perfect square factor greater than 1. Instead, we can split the surds using the fact that, for any positive numbers π‘Ž and 𝑏, we have that βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘.

Let’s start with √10; we can factor 10 into primes to note that 10=2Γ—5. Thus, √10=√2Γ—5=√2Γ—βˆš5.

Next, we can rewrite √15 in the same way, by noting that 15=5Γ—3. Therefore, √15=√5Γ—3=√5Γ—βˆš3.

Substituting these expressions into the original product yields √10Γ—βˆš15=√2Γ—βˆš5Γ—βˆš5Γ—βˆš3.

We can simplify by noting that √5Γ—βˆš5=5 and √2Γ—βˆš3=√2Γ—3=√6.

Hence, √2Γ—βˆš5Γ—βˆš5Γ—βˆš3=5√2Γ—3=5√6.

We cannot simplify any further since 6 has no perfect square factors greater than 1.

In our next example, we will simplify the quotient of two surds.

Example 3: Simplifying a Quotient of Surds

Simplify √42√21.

Answer

To simplify the quotient of two surds, we first recall that, for any positive numbers π‘Ž and 𝑏, we have that ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘. Substituting π‘Ž=42 and 𝑏=21 into this result gives us √42√21=ο„ž4221.

We note that 42÷21=2; therefore, our expression simplifies to √2.

In our next example, we will simplify an expression involving three terms that are surds.

Example 4: Simplifying the Addition and Subtraction of Three Surds with Different Radicands

Simplify √12βˆ’2√3+4√27.

Answer

To simplify this expression, we can start by simplifying each term separately. We need to write each surd in the form π‘βˆšπ‘ž, where π‘ž has no perfect square factors greater than 1. We can do this by recalling that, for any positive numbers π‘Ž and 𝑏, we have that βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘.

Let’s start by simplifying √12: we factor 12 to see that 12=4Γ—3. So, we have that √12=√4Γ—3=√4Γ—βˆš3=2√3.

We cannot simplify this surd any further since 3 is a prime; it has no perfect square factors greater than 1.

The second term is βˆ’2√3; we have already noted that we cannot simplify this surd any further.

Finally, to simplify 4√27, we factor 27 into 27=9Γ—3. Thus, 4√27=4√9Γ—3=4ο€»βˆš9Γ—βˆš3=4ο€»3√3=12√3.

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

We can take out the shared factor of √3 to obtain (2βˆ’2+12)√3, which simplifies to =12√3.

In our next example, we will use this process of simplifying surds to find an unknown value in an equation.

Example 5: 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

The right-hand side of the equation is a surd written in simplified form. We can find the value of π‘₯ by rewriting the left-hand side in the same form and then comparing the two sides.

To simplify the left-hand side of the equation, we can simplify the two surds individually. To simplify a surd, we need to write it in the form π‘βˆšπ‘ž, where π‘ž has no perfect square factors greater than 1. We can do this by recalling that, for any positive numbers π‘Ž and 𝑏, we have that βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘.

First, we can factor 8 to see that 8=4Γ—2. Thus, we can simplify as follows: 9√8=9√4Γ—2=9ο€»βˆš4Γ—βˆš2=9ο€»2√2=18√2.

Similarly, we can factor 50 to find that 50=25Γ—2. Therefore, 10√50=10√25Γ—2=10ο€»βˆš25Γ—βˆš2=10ο€»5√2=50√2.

We can use these expressions to rewrite the left-hand side of the given equation: 9√8+10√50=18√2+50√2.

This simplifies to 18√2+50√2=68√2.

We now have that 68√2=π‘₯√2,

so π‘₯=68.

In our next example, we will simplify an expression involving multiplication of surds.

Example 6: Multiplying Two Radical Expressions with the Same Base

Simplify 5√7ο€»1+√7.

Answer

We cannot simplify either of the factors in the given expression since 7 is a prime number; it has no perfect square factors greater than 1. Instead, we can distribute 5√7 into the parentheses to rewrite the expression as follows: 5√7ο€»1+√7=ο€»5√7Γ—1+ο€»5√7Γ—βˆš7=5√7+5ο€»βˆš7Γ—βˆš7.

We note that √7Γ—βˆš7=√7=7, so 5√7+5ο€»βˆš7Γ—βˆš7=5√7+(5Γ—7)=5√7+35.

It is worth noting that we can apply these results even when we are multiplying surds with different radicands. For example, consider √6Γ—βˆš2. We can simplify this using the result βˆšπ‘ŽΓ—βˆšπ‘=βˆšπ‘ŽΓ—π‘, with π‘Ž=6 and 𝑏=2. We have √6Γ—βˆš2=√6Γ—2=√12.

We then note that 12=4Γ—3, so √12=√4Γ—3=√4Γ—βˆš3=2√3.

In our final example, we will multiply two expressions involving surds and then simplify the resulting expression.

Example 7: Multiplying Two Radical Expressions and Simplifying

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

Answer

To simplify this expression, we could multiply out the parentheses; however, in this case, we have the factored form of a difference of squares: (π‘Ž+𝑏)(π‘Žβˆ’π‘)=π‘Žβˆ’π‘, with π‘Ž=5√10 and 𝑏=3√3. Therefore, we can rewrite the expression as ο€»5√10+3√35√10βˆ’3√3=ο€»5√10ο‡βˆ’ο€»3√3.

We can evaluate each square separately by recalling that (π‘π‘ž)=π‘π‘žοŠ¨οŠ¨οŠ¨. Therefore, ο€»5√10=5Γ—ο€»βˆš10=25Γ—10=250,ο€»3√3=3Γ—ο€»βˆš3=9Γ—3=27.

Hence, ο€»5√10ο‡βˆ’ο€»3√3=250βˆ’27=223.

Although this is outside the scope of this lesson, it is worth noting that the reasoning and methodology that we have applied to square roots throughout this explainer can be generalized to other orders of roots (e.g., cube roots and 𝑛th roots), providing that the roots have the same order.

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

Key Points

  • For any positive numbers π‘Ž and 𝑏, we have that
    • βˆšπ‘Žπ‘=βˆšπ‘ŽΓ—βˆšπ‘,
    • ο„žπ‘Žπ‘=βˆšπ‘Žβˆšπ‘.
  • We can use these properties to rewrite surds in the simplified form π‘βˆšπ‘ž, where π‘ž has no perfect square factors greater than 1.
  • When simplifying an expression of the form π‘βˆšπ‘ž, it is a good idea to look for the largest perfect square factor of π‘ž, as this will simplify the calculations. If this is difficult, we can always take the approach of prime factoring π‘ž as a way of determining its perfect square factors.

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