# P.3: Review - Radicals (Square Roots) - Mathematics

## Evaluating Square Roots

When the square root of a number is squared, the result is the original number. Since (4^2=16), the square root of (16) is (4).The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.

In general terms, if (a) is a positive real number, then the square root of (a) is a number that, when multiplied by itself, gives (a). The square root could be positive or negative because multiplying two negative numbers gives a positive number. However, the ( sqrt{ ext{ } } ) symbol denotes just the non-negative result, or so-called principal square root. The square root obtained using a calculator is the principal square root.

The principal square root is the nonnegative number that when multiplied by itself equals (a). The principal square root of (a) is written as (sqrt{a}). The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression. Example (PageIndex{1})

Does (sqrt{25} = pm 5)?

Solution

No. Although both (5^2) and ((−5)^2) are (25), the radical symbol implies only a nonnegative root, the principal square root. The principal square root of (25) is (sqrt{25}=5).

Note

The principal square root of (a) is the nonnegative number that, when multiplied by itself, equals (a). It is written as a radical expression (sqrt{a}), with the symbol called a radical, over the term (a), called the radicand. (sqrt{a}).

Example (PageIndex{2}): Evaluating Square Roots

Evaluate each expression.

1. (sqrt{100})
2. (sqrt{sqrt{16}})
3. (sqrt{25+144})
4. (sqrt{49})-(sqrt{81})

Solution

1. (sqrt{100} =10) because (10^2=100)
2. (sqrt{sqrt{16}}= sqrt{4} =2) because (4^2=16) and (2^2=4)
3. (sqrt{25+144} = sqrt{169} =13) because (13^2=169)
4. (sqrt{49} -sqrt{81} =7−9 =−2) because (7^2=49) and (9^2=81)

Example (PageIndex{3}):

For (sqrt{25+144}), can we find the square roots before adding?

Solution

No. (sqrt{25} + sqrt{144} =5+12=17). This is not equivalent to (sqrt{25+144}=13). The order of operations requires us to add the terms of the radicand together before finding the square root.

Try It (PageIndex{3})

Evaluate each expression.

 a. (sqrt{25}) b. (sqrt{sqrt{81}}) c. (sqrt{25-9}) d. (sqrt{36} + sqrt{121})

## Using the Product Rule to Simplify Square Roots

To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite (sqrt{15}) as (sqrt{3} imessqrt{5}). We can also use the product rule to express the product of multiple radical expressions as a single radical expression.

The Product Rule For Simplifying Square Roots

If (a) and (b) are nonnegative, the square root of the product (ab) is equal to the product of the square roots of (a) and (b)

[sqrt{ab}=sqrt{a} imessqrt{b}]

HOWTO: Simplify the square root of a product.

1. Factor any perfect squares from the radicand.
3. Simplify.

Example (PageIndex{4}): Using the Product Rule to Simplify Square Roots

1. (sqrt{300})
2. (sqrt{162a^5b^4})

Solution

( sqrt{100} imessqrt{3} ) Write radical expression as product of radical expressions.

( 10sqrt{3}) Simplify

(9a^2b^2sqrt{2a}) Simplify

Try It (PageIndex{4})

Simplify (sqrt{50x^2y^3z})

(5|x||y|sqrt{2yz})

Notice the absolute value signs around (x) and (y)? That’s because their value must be positive!

Howto: Simplify the product of multiple radical expressions

1. Express the product of multiple radical expressions as a single radical expression.
2. Simplify.

Example (PageIndex{5}): Using the Product Rule to Simplify the Product of Multiple Square Roots

a. (sqrt{12} imessqrt{3})

b. (sqrt { 6 x ^ { 3 } y ^ { 3 } } imes sqrt{2 x^3} ).

Solution

a. ( quad ext{Express the product as a single radical expression: } sqrt{12 imes3} = sqrt{36} = 6 )

b. ( quad )Begin by writing as a single radical expression: (sqrt { 12 x ^ { 6 } y ^ { 3 } }).

Determine the square factors of (12, x^{ 6}) , and (y^{ 3}).

(left. egin{array} { l } { 12 = color{Cerulean}{2 ^ { 2} }color{black}{ cdot} 3 } { x ^ { 6 } = color{Cerulean}{left( x ^ { 3 } ight) ^ { 2 } }} { y ^ { 3 } = color{Cerulean}{y ^ { 2} }color{black}{ cdot} y } end{array} ight} quadcolor{Cerulean}{Square:factors})

Make these substitutions, and then apply the product rule for radicals and simplify.

(egin{aligned} sqrt { 12 x ^ { 6 } y ^ { 3 } } & = sqrt { color{Cerulean}{2 ^ { 2} }color{black}{ cdot} 3 cdot color{Cerulean}{left( x ^ { 3 } ight) ^ { 2 }}color{black}{ cdot}color{Cerulean}{ y ^ { 2} }color{black}{ cdot} y } quadquadquadquad:color{Cerulean} { Apply:the: product :rule: for: radicals. } & = sqrt { 2 ^ { 2 } } cdot sqrt { left( x ^ { 3 } ight) ^ { 2 } } cdot sqrt { y ^ { 2 } } cdot sqrt { 3 y } quadcolor{Cerulean} { Simplify. } & = 2 cdot x ^ { 3 } cdot | y | cdot sqrt { 3 y } & = 2 x ^ { 3 } | y | sqrt { 3 y } end{aligned})

Notice the absolute value bars around ( y ). That is because the result of taking a square root is never negative.

Try It (PageIndex{5})

Simplify (sqrt{50x} imessqrt{2x}) assuming (x>0).

(10|x|)

## Using the Quotient Rule to Simplify Square Roots

Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots.

THE QUOTIENT RULE FOR SIMPLIFYING SQUARE ROOTS

The square root of the quotient (dfrac{a}{b}) is equal to the quotient of the square roots of (a) and (b) , where (a ge 0) and (b > 0) .

[sqrt{dfrac{a}{b}}=dfrac{sqrt{a}}{sqrt{b}}]

Howto: Simplify the radical of a quotient

2. Write the simplified quotient radical expression as the quotient of two radical expressions.
3. Simplify the numerator and denominator.

Example (PageIndex{6}): Using the Quotient Rule to Simplify Square Roots

 a. (sqrt{dfrac{5}{36}}) b. (sqrt { dfrac { 18 a ^ { 5 } } { b ^ { 8 } } })

Solution

a. Write as a quotient of two radical expressions and then simplify the denominator: (dfrac{sqrt{5}}{sqrt{36} = dfrac{sqrt{5}}{6})

b. Begin by determining the square factors of (18), (a^{5}), and (b^{8}).

(left. egin{array} { l } { 18 = 2 cdot color{Cerulean}{3 ^ { 2} } } { a ^ { 5 } = a ^ { 2 } cdot a ^ { 2 } cdot a = color{Cerulean}{left( a ^ { 2 } ight) ^ { 2 }}color{black}{ cdot} a } { b ^ { 8 } = b ^ { 4 } cdot b ^ { 4 } quad::= color{Cerulean}{left( b ^ { 4 } ight) ^ { 2 }} } end{array}quad ight}quadcolor{Cerulean}{Square:factors})

( quad ) Make these substitutions, apply the product and quotient rules for radicals, and then simplify.

(egin{aligned} sqrt { frac { 18 a ^ { 5 } } { b ^ { 8 } } } & = sqrt { frac { 2 cdot color{Cerulean}{3 ^ { 2} }color{black}{ cdot}color{Cerulean}{ left( a ^ { 2 } ight) ^ { 2 }}color{black}{ cdot} a } { color{Cerulean}{left( b ^ { 4 } ight) ^ { 2 } } }}quadquadcolor{Cerulean}{Apply:the:product:and:quotient:rule:for:radicals.} & = frac { sqrt { 3 ^ { 2 } } cdot sqrt { left( a ^ { 2 } ight) ^ { 2 } } cdot sqrt { 2 a } } { sqrt { left( b ^ { 4 } ight) ^ { 2 } } }quad color{Cerulean}{Simplify.} & = frac { 3 a ^ { 2 } sqrt { 2 a } } { b ^ { 4 } } end{aligned})

Try It (PageIndex{6})

Simplify (sqrt{dfrac{2x^2}{9y^4}})

(dfrac{|x|sqrt{2}}{3y^2} quad ) We do not need the absolute value signs for (y^2) because that term will always be nonnegative.

Howto: Simplify the quotient of multiple radical expressions

1. Express the quotient of multiple radical expressions as a quotient of a single radical expression.

Example (PageIndex{7}): Use the Quotient Rule to Simplify the Quotient of Two Square Roots

 a. (dfrac{sqrt{234x^{11}y}}{sqrt{26x^7y}}) b. ( dfrac { sqrt {50 x^6 y^4} } { sqrt {8 x^3 y} } )

Solution

a. First combine numerator and denominator into one radical expression

(egin{align*}
dfrac { sqrt { 50 x ^ { 6 } y ^ { 4} } } { sqrt { 8 x ^ { 3 } y } }&=sqrt{dfrac{234x^{11}y}{26x^7y}} & & ext{Write under one radical}[5pt]
&=3x^2 & & ext{Simplify square root}
end{align*})

b. Write as a single square root and cancel common factors before simplifying.

(egin{aligned} frac { sqrt { 50 x ^ { 6 } y ^ { 4 } } } { sqrt { 8 x ^ { 3 } y } } & = sqrt { frac { 50 x ^ { 6 } y ^ { 4 } } { 8 x ^ { 3 } y } } quadcolor{Cerulean}{Apply:the:quotient:rule:for:radicals:and:cancel.} & = sqrt { frac { 25 x ^ { 3 } y ^ { 3 } } { 4 } } quadcolor{Cerulean}{Simplify.} & = frac { sqrt { 25 x ^ { 3 } y ^ { 3 } } } { sqrt { 4 } } & = frac { 5 |x y| sqrt { x y } } { 2 } end{aligned})

Try It (PageIndex{7})

Simplify (dfrac{sqrt{9a^5b^{14}}}{sqrt{3a^4b^5}})

(b^4sqrt{3ab})

From this point on the assumption will be made that all variables represent non-negative real numbers. Therefore using absolute values when simplifying will not be necessary.

## Adding and Subtracting Square Roots

We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of (sqrt{2}) and (3sqrt{2}) is (4sqrt{2}). However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression (sqrt{18}) can be written with a (2) in the radicand, as (3sqrt{2}), so (sqrt{2}+sqrt{18}=sqrt{2}+3sqrt{2}=4sqrt{2})

Howto: Simplify a radical expression requiring addition or subtraction of square roots

Often, we will have to simplify before we can identify the like radicals within the terms. If the radicand and the index are not exactly the same, then the radicals are not similar and we cannot combine them.

Example​ (PageIndex{8}): Add or Subtract Square Roots

Perform the indicated operation and simplify.

1. (4 sqrt { 10 } - 5 sqrt { 10 } )
2. (5sqrt{12}+2sqrt{3}).
3. (20sqrt{72a^3b^4c}-14sqrt{8a^3b^4c})

Solution

a. (egin{aligned} ext{ }& ext{ } & ext{ } && = ( 4 - 5 ) sqrt { 10 } & & &= - 1 sqrt { 10 } & & &= - sqrt { 10 } end{aligned})

b. ( quad )We can rewrite (5sqrt{12}) as (5sqrt{4 imes3}). According the product rule, this becomes (5sqrt{4}sqrt{3}). The square root of (sqrt{4}) is (2) , so the expression becomes (5 imes2sqrt{3}), which is (10sqrt{3}). Now we can the terms have the same radicand so we can add.

[10sqrt{3}+2sqrt{3}=12sqrt{3} onumber]

[egin{align*} 20sqrt{72a^3b^4c} &= 20sqrt{9}sqrt{4}sqrt{2}sqrt{a}sqrt{a^2}sqrt{(b^2)^2}sqrt{c} &= 20(3)(2)|a|b^2sqrt{2ac} &= 120|a|b^2sqrt{2ac} end{align*}]

[egin{align*} 14sqrt{8a^3b^4c} &= 14sqrt{2}sqrt{4}sqrt{a}sqrt{a^2}sqrt{(b^2)^2}sqrt{c} &= 14(2)|a|b^2sqrt{2ac} &= 28|a|b^2sqrt{2ac} end{align*}]

Now the terms have the same radicand so we can subtract.

[120|a|b^2sqrt{2ac}-28|a|b^2sqrt{2ac}=92|a|b^2sqrt{2ac}]

Try It (PageIndex{8})

2. Subtract (3sqrt{80x}-4sqrt{45x})

a. (13sqrt{5}) ( qquad ) b. (0)

Example (PageIndex{9}): Adding and Subtracting Square Roots

Simplify.

1. ( 10 sqrt { 5 } + 6 sqrt { 2 } - 9 sqrt { 5 } - 7 sqrt { 2 } )
2. ( sqrt { 32 } - sqrt { 18 } + sqrt { 50 } )
3. (2 a sqrt { 125 a ^ { 2 } b } - a ^ { 2 } sqrt { 80 b } + 4 sqrt { 20 a ^ { 4 } b }).

Solution

a.

(egin{aligned} & ext{ } && = color{Cerulean}{10 sqrt { 5 } - 9 sqrt { 5 }}color{black}{ +}color{OliveGreen}{ 6 sqrt { 2 } - 7 sqrt { 2 }} &&& = sqrt { 5 } - sqrt { 2 } end{aligned})

We cannot simplify any further because (sqrt{5}) and (sqrt{2}) are not like radicals; the radicands are not the same.
(color{YellowOrange}{ ext{Caution:}}) It is important to point out that (sqrt { 5 } - sqrt { 2 } eq sqrt { 5 - 2 }). We can verify this by calculating the value of each side with a calculator.

( sqrt { 5 } - sqrt { 2 } approx 0.82 qquad ext{ is not the same as } qquad sqrt { 5 - 2 } = sqrt { 3 } approx 1.73 )

b.

(egin{aligned} & ext{ } && = sqrt { 16 } cdot 2 - sqrt { 9 cdot 2 } + sqrt { 25 cdot 2 } &&& = 4 sqrt { 2 } - 3 sqrt { 2 } + 5 sqrt { 2 } &&& = 6 sqrt { 2 } end{aligned})

At first glance, the radicals do not appear to be similar. However, after simplifying completely, we see that we can combine them.

c. ( ;) Step 1: Simplify the radical expression.
( quad ) Step 2: Combine all like radicals. Remember to add only the coefficients; the variable parts remain the same.

Try It (PageIndex{9})

Simplify: (sqrt { 20 } + sqrt { 27 } - 3 sqrt { 5 } - 2 sqrt { 12 }).

(- sqrt { 5 } - sqrt { 3 })

Note

Take careful note of the differences between products and sums within a radical. Assume both (x) and (y) are nonnegative.

(egin{array} { l } Products quad quadquadquad Sumshline { sqrt { x ^ { 2 } y ^ { 2 } } = x y quadsqrt { x ^ { 2 } + y ^ { 2 } } eq x + y } end{array})

The property says that we can simplify radicals when the operation in the radicand is multiplication. There is no corresponding property for addition.

## Simplifying Products of Expressions containing Square Roots

Often, there will be coefficients in front of the radicals.

Example (PageIndex{10.1x}):

Multiply: (3 sqrt { 6 } cdot 5 sqrt { 2 })

Solution

Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows.

(egin{aligned} 3 sqrt { 6 } cdot 5 sqrt { 2 } & = color{Cerulean}{3 cdot 5}color{black}{ cdot}color{OliveGreen}{ sqrt { 6 } cdot sqrt { 2} }quadcolor{Cerulean}{Multiplication:is:commutative.} & = 15 cdot sqrt { 12 } quadquadquad:color{Cerulean}{Multiply:the:coefficients:and:the:radicands.} & = 15 sqrt { 4 cdot 3 } quadquadquad:color{Cerulean}{Simplify.} & = 15 cdot 2 cdot sqrt { 3 } & = 30 sqrt { 3 } end{aligned})

Typically, the first step involving the application of the commutative property is not shown.

We will often find the need to subtract a radical expression with multiple terms. If this is the case, remember to apply the distributive property before combining like terms.

Example (PageIndex{10.2x}): Use the distributive property with square roots

(egin{aligned} & ext{Simplify: } ( 5 sqrt { x } - 4 sqrt { y } ) - ( 4 sqrt { x } - 7 sqrt { y } ) && ext{Solution:} ext{ }& ext{ } & ext{ } && = 5 sqrt { x } - 4 sqrt { y } - 4 sqrt { x } + 7 sqrt { y } quadcolor{Cerulean}{Distribute.} &&& = 5 sqrt { x } - 4 sqrt { x } - 4 sqrt { y } + 7 sqrt { y } &&& = sqrt { x } + 3 sqrt { y } end{aligned})

Use the distributive property when multiplying rational expressions with more than one term.

Example (PageIndex{10.3x}):

Multiply: (5 sqrt { 2 x } ( 3 sqrt { x } - sqrt { 2 x } )).

Solution:

Apply the distributive property and multiply each term by (5 sqrt { 2 x }).

(egin{aligned} 5 sqrt { 2 x } ( 3 sqrt { x } - sqrt { 2 x } ) & = color{Cerulean}{5 sqrt { 2 x } }color{black}{cdot} 3 sqrt { x } - color{Cerulean}{5 sqrt { 2 x }}color{black}{ cdot} sqrt { 2 x } quadcolor{Cerulean}{Distribute.} & = 15 sqrt { 2 x ^ { 2 } } - 5 sqrt { 4 x ^ { 2 } } quadquadquadquad:::color{Cerulean}{Simplify.} & = 15 x sqrt { 2 } - 5 cdot 2 x & = 15 x sqrt { 2 } - 10 x end{aligned})

The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.

Example (PageIndex{10.4x}):

Multiply: (( sqrt { x } - 5 sqrt { y } ) ^ { 2 }).

Solution

(( sqrt { x } - 5 sqrt { y } ) ^ { 2 } = ( sqrt { x } - 5 sqrt { y } ) ( sqrt { x } - 5 sqrt { y } ))

Begin by applying the distributive property. (egin{array} { l } { = color{Cerulean}{sqrt { x }}color{black}{ cdot} sqrt { x } + color{Cerulean}{sqrt { x }}color{black}{ (} - 5 sqrt { y } ) + ( color{OliveGreen}{- 5 sqrt { y }}color{black}{ )} sqrt { x } + ( color{OliveGreen}{- 5 sqrt { y }}color{black}{ )} ( - 5 sqrt { y } ) } { = sqrt { x ^ { 2 } } - 5 sqrt { x y } - 5 sqrt { x y } + 25 sqrt { y ^ { 2 } } } { = x - 10 sqrt { x y } + 25 y } end{array})

The binomials ((a + b)) and ((a − b)) are called conjugates. When multiplying conjugate binomials the middle terms are opposites and their sum is zero.

Example (PageIndex{10.5x}):

Multiply: (( sqrt { 10 } + sqrt { 3 } ) ( sqrt { 10 } - sqrt { 3 } )).

Solution

Apply the distributive property, and then combine like terms.

(egin{aligned} ( sqrt { 10 } + sqrt { 3 } ) ( sqrt { 10 } - sqrt { 3 } ) & = color{Cerulean}{sqrt { 10} }color{black}{ cdot} sqrt { 10 } + color{Cerulean}{sqrt { 10} }color{black}{ (} - sqrt { 3 } ) + color{OliveGreen}{sqrt{3}}color{black}{ (}sqrt{10}) + color{OliveGreen}{sqrt{3}}color{black}{(}-sqrt{3}) & = sqrt { 100 } - sqrt { 30 } + sqrt { 30 } - sqrt { 9 } & = 10 - color{red}{sqrt { 30 }}color{black}{ +}color{red}{ sqrt { 30} }color{black}{ -} 3 & = 10 - 3 & = 7 end{aligned})

It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general

(egin{aligned} ( sqrt { x } + sqrt { y } ) ( sqrt { x } - sqrt { y } ) & = sqrt { x ^ { 2 } } - sqrt { x y } + sqrt {x y } - sqrt { y ^ { 2 } } & = x - y end{aligned})

Alternatively, using the formula for the difference of squares we have,

(egin{aligned} ( a + b ) ( a - b ) & = a ^ { 2 } - b ^ { 2 }quadquadquadcolor{Cerulean}{Difference:of:squares.} ( sqrt { x } + sqrt { y } ) ( sqrt { x } - sqrt { y } ) & = ( sqrt { x } ) ^ { 2 } - ( sqrt { y } ) ^ { 2 } & = x - y end{aligned})

Try It (PageIndex{10x})

Multiply: (( 3 - 2 sqrt { y } ) ( 3 + 2 sqrt { y } )). (Assume (y) is positive).

(9-4y)

## Rationalizing Denominators

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.

We know that multiplying by (1) does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of (1) that will eliminate the radical.

### Monomial Denominators

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is (bsqrt{c}), multiply by (dfrac{sqrt{c}}{sqrt{c}}).

HowTo: Given an expression with a single square root radical term in the denominator, rationalize the denominator

1. Multiply the numerator and denominator by the radical in the denominator.
2. Simplify.

Sometimes, we will find the need to reduce, or cancel, after rationalizing the denominator.

Example (PageIndex{11}): Rationalize a Denominator Containing a Single Term

Write in simplest form (rationalize the denominator).

 a. ( dfrac{2sqrt{3}}{3sqrt{10}} ) b. (dfrac { sqrt { 2 } } { sqrt { 5 x } }) c. (dfrac { 3 a sqrt { 2 } } { sqrt { 6 a b } })

Solution

a. The radical in the denominator is (sqrt{10}). So multiply the fraction by (dfrac{sqrt{10}}{sqrt{10}}). Then simplify.

( dfrac{2sqrt{3}}{3sqrt{10}} imes {color{Cerulean}{ dfrac{sqrt{10}}{sqrt{10}} } } = dfrac{2sqrt{30}}{30} = dfrac{sqrt{30}}{15} )

b. The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by (1) in the form (frac { sqrt { 5 x } } { sqrt { 5 x } }).

(egin{aligned} frac { sqrt { 2 } } { sqrt { 5 x } } & = frac { sqrt { 2 } } { sqrt { 5 x } } cdot color{Cerulean}{frac { sqrt { 5 x } } { sqrt { 5 x } } } & = frac { sqrt { 10 x } } { sqrt { 25 x ^ { 2 } } } quadquad: color{Cerulean} { Simplify. } & = frac { sqrt { 10 x } } { 5 x } end{aligned})

c. In this example, we will multiply by (1) in the form (frac { sqrt { 6 a b } } { sqrt { 6 a b } }).

(egin{aligned} frac { 3 a sqrt { 2 } } { sqrt { 6 a b } } & = frac { 3 a sqrt { 2 } } { sqrt { 6 a b } } cdot color{Cerulean}{frac { sqrt { 6 a b } } { sqrt { 6 a b } }} & = frac { 3 a sqrt { 12 a b } } { sqrt { 36 a ^ { 2 } b ^ { 2 } } } quadquadcolor{Cerulean}{Simplify.} & = frac { 3 a sqrt { 4 cdot 3 a b} } { 6 ab } & = frac { 6 a sqrt { 3 a b } } { b }quadquad::color{Cerulean}{Cancel.} & = frac { sqrt { 3 a b } } { b } end{aligned})

Notice that (b) does not cancel in this example. Do not cancel factors inside a radical with those that are outside.

Try It (PageIndex{11})

Write in simplest form (rationalize the denominator).

 a. (dfrac{12sqrt{3}}{sqrt{2}}) b. (sqrt { dfrac { 9 x } { 2 y } })

a. (6sqrt{6}) ( qquad ) b. (frac { 3 sqrt { 2xy } } { 2 y })

### Binomial Denominators

For a denominator containing the sum or difference of rational or irrational terms, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign that connects the two terms in the denominator. If the denominator is (a+bsqrt{c}) , then the conjugate is (a-bsqrt{c}).

How to: Given an expression with a radical term and a constant in the denominator, rationalize the denominator

1. Find the conjugate of the denominator.
2. Multiply the numerator and denominator by the conjugate.
3. Use the distributive property.
4. Simplify.

Example (PageIndex{12}): Rationalizing a Denominator Containing Two Terms

Write in simplest form (rationalize the denominator).

 a. (dfrac{4}{1+sqrt{5}}) b. (dfrac { 1 } { sqrt { 5 } - sqrt { 3 } }) c. (dfrac { sqrt { 10 } } { sqrt { 2 } + sqrt { 6 } }) d. (dfrac { sqrt { x } - sqrt { y } } { sqrt { x } + sqrt { y } })

Solution

a. Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of (1+sqrt{5}) is (1-sqrt{5}). Then multiply the fraction by ( frac{1-sqrt{5}}{1-sqrt{5}}) .

[egin{align*} &dfrac{4}{1+sqrt{5}} imes { color{Cerulean} { dfrac{1-sqrt{5}}{1-sqrt{5}} } }[5pt] &dfrac{4-4sqrt{5}}{-4} & & ext{Use the distributive property}[5pt] &-1 + sqrt{5} & & ext{Simplify} end{align*}]

b. In this example, the conjugate of the denominator is (sqrt { 5 } + sqrt { 3 }). Therefore, multiply by (1) in the form (frac { ( sqrt { 5 } + sqrt { 3 } ) } { ( sqrt {5 } + sqrt { 3 } ) }).

(egin{aligned} frac { 1 } { sqrt { 5 } - sqrt { 3 } } & = frac { 1 } { ( sqrt { 5 } - sqrt { 3 } ) } color{Cerulean}{frac { ( sqrt { 5 } + sqrt { 3 } ) } { ( sqrt { 5 } + sqrt { 3 } ) } ::Multiply :by:"1".} & = frac { sqrt { 5 } + sqrt { 3 } } { sqrt { 25 } + sqrt { 15 } - sqrt{15}-sqrt{9} } :color{Cerulean}{Simplify.} & = frac { sqrt { 5 } + sqrt { 3 } } { 5-3 } & = frac { sqrt { 5 } + sqrt { 3 } } { 2 } end{aligned})

Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property (( sqrt { a } + sqrt { b } ) ( sqrt { a } - sqrt { b } ) = a - b) to expedite the process of multiplying the expressions in the denominator.

c. Multiply by (1) in the form (frac { sqrt { 2 } - sqrt { 6 } } { sqrt { 2 } - sqrt { 6 } }).

(egin{aligned} frac{sqrt{10}}{sqrt{2}+sqrt{6} }&= frac{(sqrt{10})}{(sqrt{2}+sqrt{6})} color{Cerulean}{frac{(sqrt{2}-sqrt{6})}{(sqrt{2}-sqrt{6})}quadquad Multiple:by:the:conjugate.} &= frac { sqrt { 20 } - sqrt { 60 } } { 2 - 6 } quadquadquadquadquadquad:::color{Cerulean}{Simplify.} &= frac { sqrt { 4 cdot 5 } - sqrt { 4 cdot 15 } } { - 4 } &= frac { 2 sqrt { 5 } - 2 sqrt { 15 } } { - 4 } &=frac{2(sqrt{5}-sqrt{15})}{-4} &= frac { sqrt { 5 } - sqrt { 15 } } { - 2 } = - frac { sqrt { 5 } - sqrt { 15 } } { 2 } = frac { - sqrt { 5 } + sqrt { 15 } } { 2 } end{aligned})

d. In this example, we will multiply by (1) in the form (frac { sqrt { x } - sqrt { y } } { sqrt { x } - sqrt { y } }).

(egin{aligned} frac { sqrt { x } - sqrt { y } } { sqrt { x } + sqrt { y } } & = frac { ( sqrt { x } - sqrt { y } ) } { ( sqrt { x } + sqrt { y } ) } color{Cerulean}{frac { ( sqrt { x } - sqrt { y } ) } { ( sqrt { x } - sqrt { y } ) } quad quad Multiply:by:the:conjugate:of:the:denominator.} & = frac { sqrt { x ^ { 2 } } - sqrt { x y } - sqrt { x y } + sqrt { y ^ { 2 } } } { x - y } ::color{Cerulean}{Simplify.} & = frac { x - 2 sqrt { x y } + y } { x - y } end{aligned})

Try It (PageIndex{12})

Write in simplest form (rationalize the denominator).

 a. (dfrac{7}{2+sqrt{3}}) b. (dfrac { 2 sqrt { 3 } } { 5 - sqrt { 3 } })

a. (14-7sqrt{3}) ( qquad ) b. (dfrac { 5 sqrt { 3 } + 3 } { 11 })

## 9.3 Add and Subtract Square Roots

We know that we must follow the order of operations to simplify expressions with square roots. The radical is a grouping symbol, so we work inside the radical first. We simplify 2 + 7 2 + 7 in this way:

But, just like we can add x + x , we can add 3 + 3 . x + x = 2 x 3 + 3 = 2 3 But, just like we can add x + x , we can add 3 + 3 . x + x = 2 x 3 + 3 = 2 3

Adding square roots with the same radicand is just like adding like terms. We call square roots with the same radicand like square roots to remind us they work the same as like terms.

### Like Square Roots

Square roots with the same radicand are called like square roots .

We add and subtract like square roots in the same way we add and subtract like terms. We know that 3 x + 8 x 3 x + 8 x is 11 x 11 x . Similarly we add 3 x + 8 x 3 x + 8 x and the result is 11 x . 11 x .

### Add and Subtract Like Square Roots

Think about adding like terms with variables as you do the next few examples. When you have like radicands, you just add or subtract the coefficients. When the radicands are not like, you cannot combine the terms.

### Example 9.32

Simplify: 5 13 + 4 13 + 2 13 5 13 + 4 13 + 2 13 .

#### Solution

Simplify: 4 11 + 2 11 + 3 11 4 11 + 2 11 + 3 11 .

Simplify: 6 10 + 2 10 + 3 10 6 10 + 2 10 + 3 10 .

### Example 9.34

Simplify: 2 5 n − 6 5 n + 4 5 n 2 5 n − 6 5 n + 4 5 n .

#### Solution

Simplify: 7 x − 7 7 x + 4 7 x 7 x − 7 7 x + 4 7 x .

Simplify: 4 3 y − 7 3 y + 2 3 y 4 3 y − 7 3 y + 2 3 y .

When radicals contain more than one variable, as long as all the variables and their exponents are identical, the radicals are like.

### Example 9.35

Simplify: 3 x y + 5 3 x y − 4 3 x y 3 x y + 5 3 x y − 4 3 x y .

#### Solution

Simplify: 5 x y + 4 5 x y − 7 5 x y 5 x y + 4 5 x y − 7 5 x y .

Simplify: 3 7 m n + 7 m n − 4 7 m n 3 7 m n + 7 m n − 4 7 m n .

### Add and Subtract Square Roots that Need Simplification

Remember that we always simplify square roots by removing the largest perfect-square factor. Sometimes when we have to add or subtract square roots that do not appear to have like radicals , we find like radicals after simplifying the square roots.

### Example 9.39

Simplify: 3 4 192 − 5 6 108 3 4 192 − 5 6 108 .

#### Solution

Simplify: 2 3 108 − 5 7 147 2 3 108 − 5 7 147 .

Simplify: 3 5 200 − 3 4 128 3 5 200 − 3 4 128 .

### Example 9.40

#### Solution

In the next example, we will remove constant and variable factors from the square roots.

### Example 9.42

Simplify: 9 50 m 2 − 6 48 m 2 9 50 m 2 − 6 48 m 2 .

#### Solution

Simplify: 5 32 x 2 − 3 48 x 2 5 32 x 2 − 3 48 x 2 .

Simplify: 7 48 y 2 − 4 72 y 2 7 48 y 2 − 4 72 y 2 .

### Example 9.43

Simplify: 2 8 x 2 − 5 x 32 + 5 18 x 2 2 8 x 2 − 5 x 32 + 5 18 x 2 .

#### Solution

Simplify: 3 12 x 2 − 2 x 48 + 4 27 x 2 3 12 x 2 − 2 x 48 + 4 27 x 2 .

Simplify: 3 18 x 2 − 6 x 32 + 2 50 x 2 3 18 x 2 − 6 x 32 + 2 50 x 2 .

### Media

Access this online resource for additional instruction and practice with the adding and subtracting square roots.

### Section 9.3 Exercises

#### Practice Makes Perfect

Add and Subtract Like Square Roots

In the following exercises, simplify.

8 11 c d + 5 11 c d − 9 11 c d 8 11 c d + 5 11 c d − 9 11 c d

11 2 r s − 9 2 r s + 3 2 r s 11 2 r s − 9 2 r s + 3 2 r s

Add and Subtract Square Roots that Need Simplification

In the following exercises, simplify.

3 20 x 2 − 4 45 x 2 + 5 x 80 3 20 x 2 − 4 45 x 2 + 5 x 80

3 128 y 2 + 4 y 162 − 8 98 y 2 3 128 y 2 + 4 y 162 − 8 98 y 2

3 75 y 2 + 8 y 48 − 300 y 2 3 75 y 2 + 8 y 48 − 300 y 2

Mixed Practice

#### Everyday Math

A decorator decides to use square tiles as an accent strip in the design of a new shower, but she wants to rotate the tiles to look like diamonds. She will use 9 large tiles that measure 8 inches on a side and 8 small tiles that measure 2 inches on a side. Determine the width of the accent strip by simplifying the expression 9 ( 8 2 ) + 8 ( 2 2 ) 9 ( 8 2 ) + 8 ( 2 2 ) . (Round to the nearest tenth of an inch.)

Suzy wants to use square tiles on the border of a spa she is installing in her backyard. She will use large tiles that have area of 12 square inches, medium tiles that have area of 8 square inches, and small tiles that have area of 4 square inches. Once section of the border will require 4 large tiles, 8 medium tiles, and 10 small tiles to cover the width of the wall. Simplify the expression 4 12 + 8 8 + 10 4 4 12 + 8 8 + 10 4 to determine the width of the wall. (Round to the nearest tenth of an inch.)

#### Writing Exercises

Explain the process for determining whether two radicals are like or unlike. Make sure your answer makes sense for radicals containing both numbers and variables.

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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## Open Resources for Community College Algebra

### Subsection 6.5.1 Square and (n)th Root Properties

In Section 6.1 we defined the square root (sqrt) and (n)th root (sqrt[n]) radicals. When (x) is positive, the expression (sqrt[n]) means a positive number (r ext<,>) where (overbrace^>=x ext<.>) The square root (sqrt) is just the case where (n=2 ext<.>)

When (x) is negative, (sqrt[n]) might not be defined. It depends on whether or not (n) is an even number. When (x) is negative and (n) is odd, (sqrt[n]) is a negative number where (overbrace^>=x ext<.>)

List 6.5.1. Rules of Radicals for Multiplication and Division

If (a) and (b) are positive real numbers, and (m) is a positive integer , then we have the following rules:

### Subsection 6.5.2 Rationalizing the Denominator

In Section 6.2 we covered how to rationalize the denominator when it contains a single square root or a binomial with a square root term.

###### Example 6.5.3 .

Rationalize the denominator of the expressions.

First we will simplify (sqrt<75> ext<.>)

Now we can rationalize the denominator by multiplying the numerator and denominator by (sqrt<3> ext<.>)

###### Example 6.5.4 . Rationalize Denominator Using the Difference of Squares Formula.

Rationalize the denominator in (frac-sqrt<5>>+sqrt<2>> ext<.>)

To remove radicals in (sqrt<3>+sqrt<2>) with the difference of squares formula, we multiply it with (sqrt<3>-sqrt<2> ext<.>)

### Subsection 6.5.3 Radical Expressions and Rational Exponents

###### Example 6.5.5 . Radical Expressions and Rational Exponents.

Simplify the expressions using Fact 6.3.2 or Fact 6.3.9.

In this problem the negative can be associated with either the numerator or the denominator, but not both. We choose the numerator.

###### Example 6.5.6 . More Expressions with Rational Exponents.

Use exponent properties in List 6.3.14 to simplify the expressions, and write all final versions using radicals.

This was going to be a comment, but it got kind of long so I'm putting it here:

A big "gotcha' worth acknowledging is $sqrt$ has no "positive" option once we allow complex inputs. The ability to take the positive root is an artifact of the real numbers. We obviously want $sqrt$ to be continuous, so you might try to just extend the positive branch of $sqrt$ to the complex numbers.

The unfortunate fact is there is no way to do this. The $sqrt$ function is (in a way that complex analysis makes precise) inherently discontinuous. However, we can repair the situation in a number of ways:

Restricting the domain of $sqrt$ . This is basically the same as saying "don't take square roots of negative numbers". There are lots of possible "branch cuts" that you can take, but no matter what you do you'll be left with a discontinuity, which we typically treat as an undefined region.

Changing the domain to a riemann surface. This is really the "best" solution for a number of reasons, though it is conceptually difficult.

Allowing multi-valued functions. The issue with making $sqrt$ continuous in $mathbb C$ is this: Start at $z$ , and remember $sqrt$ . Then make a lap around the origin, taking note of $sqrt$ as you go (this should be continuously, of course). When you get back to $z$ , you'll find $-sqrt$ instead of what you started with! So we can "solve this problem" by just asserting $pm sqrt$ is the output. Then when we make a lap around the origin, we start at $pm sqrt$ and we end at $mp sqrt$ , which is the same! Our problem has "gone away", but at the cost of changing what it means to be a function.

This problem really doesn't come to light until you start playing with Complex Analysis, but $sqrt$ and $log(cdot)$ are the two most famous examples of this phenomenon. Congratulations on running into it! This exact "issue" with complex numbers is responsible for tons of recent mathematics, with really beautiful geometry to show for it.

## How to Solve Square Root Problems

This article was co-authored by David Jia. David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math.

There are 14 references cited in this article, which can be found at the bottom of the page.

While the intimidating sight of a square root symbol may make the mathematically-challenged cringe, square root problems are not as hard to solve as they may first seem. Simple square root problems can often be solved as easily as basic multiplication and division problems. More complex square root problems, on the other hand, can require some work, but with the right approach, even these can be easy. Start practicing square root problems today to learn this radical new math skill!

## How to Add Square Roots  ### Video in How To Add Square Roots

#### How to Add Simplified Square Roots?

Let's look at the following example

You can only add square roots (or radicals) that have the same radicand. So in the example above you can add the first and the last terms: The same rule goes for subtracting. Consider the following example:

You can subtract square roots with the same radicand--which is the first and last terms.

### Practice Problems

##### Problem 1

Only the first and last square root have the same radicand, so you can add these two terms. ##### Problem 2

Remember--the same rule applies to subtracting square roots--the radicands must be the same.

#### How to Add Square Roots- that are not simplified

Let's look at the following example

You may immediately see the problem here: The radicands are not the same. Therefore, we can not add them at the moment. However, if we simplify the square roots first, we will be able to add them. Let's use this example problem to illustrate the general steps for adding square roots.

## ORCCA Open Resources for Community College Algebra

### Subsection 14.5.1 Introduction to Radical Functions

In Section 14.1 we covered the square root and other root functions. We learned how to find the domain of radical functions algebraically and the range graphically. We also saw the distance formula which is an application of square roots.

###### Example 14.5.1 The Square Root Function

Algebraically find the domain of the function (f) where (f(x)=sqrt<3x-1>+2) and then find the range by making a graph.

Using Fact 14.1.29 to find the function's domain, we set the radicand greater than or equal to zero and solve:

The function's domain is (left[frac<1><3>,infty ight)) in interval notation.

To find the function's range, we use technology to look at a graph of the function. The graph shows that the function's range is ([2,infty) ext<.>) The graph also verifies the function's domain is indeed (left[frac<1><3>,infty ight) ext<.>)

###### Example 14.5.3 The Distance Formula

Find the distance between ((6,-13)) and ((-4,17) ext<.>)

To calculate the distance between ((6,-13)) and ((-4,17) ext<,>) we use the distance formula. It's good practice to mark each value with the corresponding variables in the formula:

The distance between ((6,-13)) and ((-4,17)) is (10sqrt<10>) or approximately (31.62) units.

###### Example 14.5.4 The Cube Root Function

Algebraically find the domain and graphically find the range of the function (g) where (g(x)=-2sqrt-1 ext<.>)

First note that the index of the function (g) for (g(x)=-2sqrt-1) is odd. By Fact 14.1.29, the domain is ((-infty,infty) ext<.>)

To find the function's range, we use technology to graph the function. According to the graph, the function's range is also ((-infty,infty) ext<.>)

###### Example 14.5.6 Other Roots

Algebraically find the domain and graphically find the range of the function (h) where (h(x)=8-frac<5><2>sqrt<6-2x> ext<.>)

First note that the index of this function is (4 ext<,>) which is even. By Fact 14.1.29, to find the domain of this function we must set the radicand greater or equal to zero and solve.

So, the domain of (h) is ((-infty,3] ext<.>)

To find the function's range, we use technology to graph the function. By the graph, the function's range is ((-infty,8] ext<.>)

### Subsection 14.5.2 Radical Expressions and Rational Exponents

###### Example 14.5.8 Radical Expressions and Rational Exponents

Simplify the expressions using Fact 14.2.2 or Fact 14.2.10.

In this problem the negative can be associated with either the numerator or the denominator, but not both. We choose the numerator.

###### Example 14.5.9 More Expressions with Rational Exponents

Use exponent properties in List 14.2.15 to simplify the expressions, and write all final versions using radicals.

### Subsection 14.5.3 More on Rationalizing the Denominator

In Section 14.3 we covered how to rationalize the denominator when it contains a single square root or a binomial.

###### Example 14.5.10 A Review of Rationalizing the Denominator

Rationalize the denominator of the expressions.

First we will simplify (sqrt<75> ext<.>)

Now we can rationalize the denominator by multiplying the numerator and denominator by (sqrt<3> ext<.>)

###### Example 14.5.11 Rationalize Denominator with Difference of Squares Formula

Rationalize the denominator in (frac-sqrt<5>>+sqrt<2>> ext<.>)

To remove radicals in (sqrt<3>+sqrt<2>) with the difference of squares formula, we multiply it with (sqrt<3>-sqrt<2> ext<.>)

### Subsection 14.5.4 Solving Radical Equations

In Section 14.4 we covered solving equations that contain a radical. We learned about extraneous solutions and the need to check our solutions.

###### Example 14.5.12 Solving Radical Equations

Solve for (r) in (r=9+sqrt ext<.>)

We will isolate the radical first, and then square both sides.

Because we squared both sides of an equation, we must check both solutions.

It turns out (6) is an extraneous solution and (13) is a valid solution. So the equation has one solution: (13 ext<.>) The solution set is (<13> ext<.>)

###### Example 14.5.13 Solving Radical Equations that Require Squaring Twice

Solve the equation (sqrt=-1-sqrt) for (t ext<.>)

We cannot isolate two radicals, so we will simply square both sides, and later try to isolate the remaining radical.

Because we squared both sides of an equation, we must check the solution by substituting (substitute<16>) into (sqrt=-1-sqrt ext<,>) and we have:

Our solution did not check so there is no solution to this equation. The solution set is the empty set, which can be denotes (< ext< >>) or (emptyset ext<.>)

### Subsection 14.5.5 Exercises

Find the domain of the function.

Find the domain of the function.

Find the domain of the function.

Find the domain of the function.

Find the domain of the function.

Find the domain of the function.

Use technology to find the range of the function (h) defined by (h(x)=+1> ext<.>)

Use technology to find the range of the function (F) defined by (F(x)=-5> ext<.>)

If an object is dropped with no initial velocity, the time since the drop, in seconds, can be calculated by the function

where (h) is the distance the object traveled in feet. The variable (g) is the gravitational acceleration on earth, and we can round it to (32 frac) for this problem.

After seconds since the release, the object would have traveled (35) feet.

After (5) seconds since the release, the object would have traveled feet.

If an object is dropped with no initial velocity, the time since the drop, in seconds, can be calculated by the function

where (h) is the distance the object traveled in feet. The variable (g) is the gravitational acceleration on earth, and we can round it to (32 frac) for this problem.

After seconds since the release, the object would have traveled (40) feet.

After (3.6) seconds since the release, the object would have traveled feet.

Find the distance between the points ((-8,-2)) and ((57,70) ext<.>)

Find the distance between the points ((-10,-15)) and ((45,33) ext<.>)

## P.3: Review - Radicals (Square Roots) - Mathematics

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## Simplifying Radicals: Square and Cube Roots Activity: Math Message Decoder I have been a science teacher in an urban School District for 7 years. During this time I have had the opportunity to experience teaching within classrooms at all levels of performance abilities, including full inclusion, and highly advanced classes. I have taught middle school science (both the 7th and 8th grade) and at the high school level (9th to 12th grade science) including Anatomy, Astronomy, Biology, Chemistry, Environmental Science, and Physical Science I have a PA Professional Certi

This message decoder is a great way for students to practice their skills with simplifying radicals using square roots and cube roots. The solution to each puzzle is included.

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Our customer service team will review your report and will be in touch. I have been a science teacher in an urban School District for 7 years. During this time I have had the opportunity to experience teaching within classrooms at all levels of performance abilities, including full inclusion, and highly advanced classes. I have taught middle school science (both the 7th and 8th grade) and at the high school level (9th to 12th grade science) including Anatomy, Astronomy, Biology, Chemistry, Environmental Science, and Physical Science I have a PA Professional Certi

## Properties of Roots

Every positive real number has a positive nth root and the rules for operations with such roots obey the following rules:  We can convert any expression with radicals into an exponent form: Using the exponent form makes it easier to cancel out powers and roots. We will cover exponent rules in more depth in the next chapter.

Since roots are special cases of exponents, radical rules are essentially exponent rules written with a different notation, the radical symbol.