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Algebra 1 View details
4. Operating with Square Roots
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Chapter 8
4. 

Operating with Square Roots

This lesson provides a comprehensive guide to understanding and operating with square roots. It covers key concepts such as the product property, which helps in simplifying the multiplication of square roots, and the quotient property, useful for dividing square roots. Additionally, the concept of rationalization is discussed, a technique used to eliminate square roots from the denominator of a fraction. These principles are not just theoretical; they find applications in various scientific and engineering tasks, such as calculating distances, areas, and even in some algorithms in computer science. Overall, the material serves as an invaluable lesson for mastering the complexities of square roots and their practical applications in real-world scenarios.
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18 Theory slides
9 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
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Operating with Square Roots
Slide of 18
This lesson will introduce the concept of a square root and how to operate its many properties.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

The Golden Ratio

The golden ratio is the ratio of two positive numbers with the property that their ratio is the same as the ratio of their sum to the larger of the two numbers. It is represented by the Greek letter φ. Its value is given by the following expression. φ = 1+sqrt(5)/2 Not only mathematicians, but throughout history, architects and artists have used this ratio. To this day, people still paint on canvases shaped as a golden rectangle. The length l and width w of this rectangle are in the golden ratio. l/w=φ Dylan, an avid painter, sets out to buy an uncut canvas. Feeling savvy, he thinks he can cut the canvas himself in such a way that it becomes a golden rectangle. Dylan aims to cut the canvas a length l of 60 inches. What should be the width?

Dylan Canvas Golden Ratio
Discussion

Square Roots

Some numbers cannot be expressed as the ratio of two integers. These numbers have a special name.

Concept

Irrational Numbers

The set of irrational numbers is formed by all numbers that cannot be expressed as the ratio between two integers. sqrt(2), sqrt(3), sqrt(5), e, π Irrational numbers are real numbers, but they cannot be expressed as fractions. Also, the decimal expansion of irrational numbers is not repeating and non-terminating. sqrt(2) &= 1.41421356237... π &= 3.14159265359 ... In other words, a number is irrational if it is not rational. Although this number set does not have its own symbol, it is sometimes represented with a combination of other symbols.

R-Q or R \ Q

From the examples given above, sqrt(2), sqrt(3), and sqrt(5) are called the square root of 2, the square root of 3, and the square root of 5, respectively.

Concept

Square Root

A square root of a number a is a number that, when multiplied by itself, equals a. For example, 4 and - 4 are the square roots of 16. 4* 4 &= 16 [0.5em] -4 * (-4)&=16 All positive numbers have two square roots — one positive and one negative. To avoid ambiguity, when talking about the square root of a number, only the positive root, also known as its principal root, is considered. Furthermore, to denote the square root, the symbol sqrt() is used. For example, the square root of 16 is denoted as sqrt(16). sqrt(16) = 4 In the example above, the principal root of 16 is 4 because 4 multiplied by itself equals 16 and 4 is positive. When a number is a perfect square, its square roots are integers. The square roots of positive integers that are non-perfect squares are irrational numbers.

Principal Root of Perfect Squares Principal Root of Non-Perfect Squares
Perfect Square Principal Root
(Integer Number) 		Non-Perfect Square Principal Root
(Irrational Number)
1 sqrt(1)=1 2 sqrt(2)≈ 1.414213...
4 sqrt(4)=2 3 sqrt(3)≈ 1.732050...
9 sqrt(9)=3 5 sqrt(5)≈ 2.236067...
16 sqrt(16)=4 10 sqrt(10)≈ 3.162277...
25 sqrt(25)=5 20 sqrt(20)≈ 4.472135...

Extra

 Square Roots of Fractions and Decimal Numbers

Separate from whole numbers, the square roots of fractions can be calculated by taking square roots of the numerator and denominator separately. Consider the following example. 9/16=3^2/4^2 ⇒ sqrt(9/16)=3/4 The square roots of decimal numbers can be calculated by writing them in the fraction form. Then, the square roots of the numerator and denominator are calculated. Consider the following example. 0.36=36/100=6^2/10^2 [0.5em] ⇓ sqrt(36/100)=6/10=0.6

Example

Fertilizing a Garden Using the Square Root

Emily visited her grandparent's new house for a family gathering. She loves their huge backyard! Her grandpa, eager to let her explore, told her she can use some of the free space and some leftover fertilizer to make herself a little flower garden!

Grandpa says that there is enough fertilizer to cover 81 square feet. Emily wants to use this fertilzer to make a garden in the shape of a square.
What should be the length l of one side of the garden?

Hint

What is the square root of 81?

Solution
The area of Emily's garden can be found by using the formula for the area of a square. Further, since there is enough fertilizer to cover 81 square feet, this number can be substituted into the formula. A=l^2 substitute 81=l^2 The obtained equation states that l is a number whose square equals 81. This means that l multiplied by itself is 81. Therefore, l is the square root of 81. Since l represents the side length of a square, it must be positive. Because of this fact, only the principal root will be considered.
l=sqrt(81)
CalcRoot

Calculate root

l=9
For the garden's area to be 81 square feet, the side length must be 9 feet. Emily can now start gardening in full confidence!
Discussion

Product Property of Square Roots

Sometimes it is necessary to simplify a square root. The Product Property of Square Roots can be helpful when doing so.

Given two non-negative numbers a and b, the square root of their product equals the product of the square root of each number.


sqrt(ab) = sqrt(a)* sqrt(b), for a≥ 0 and b≥ 0

Proof
Let x, y, and z be three non-negative numbers such that x=sqrt(a), y=sqrt(b), and z = sqrt(ab). By the definition of a square root, each of these numbers squared is equal to its corresponding radicand. x^2 = a & (I) y^2 = b & (II) z^2 = ab & (III) Next, multiply Equation (I) by y^2. x^2=& a ⇓ x^2 y^2=& a y^2 Now, substitute Equations (II) and (III) into this equation.
x^2 y^2 = a y^2
▼
Substitute values and simplify
Substitute

y^2= b

x^2 y^2 = a b
Substitute

ab= z^2

x^2 y^2 = z^2
ProdPowII

a^m b^m = (a b)^m

(xy)^2 = z^2
RearrangeEqn

Rearrange equation

z^2 = (xy)^2
Since x, y, and z are non-negative, the final equation implies that z=xy. z^2=(xy)^2 ⇒ z=xy The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=xy ⇔ sqrt(ab)=sqrt(a)* sqrt(b)
Example

Finding the Area of a Plot

At the family gathering, Emily's aunt named Auntie Agent is gushing about her job as a real estate agent. She is bragging about a recent business deal. She purchased a new plot that is located next to two plots she also owns, as highlighted in the diagram.

Auntie Agent wants to resale her newly purchased plot in a few years. To do so, she needs to know the area of the plot. Unfortunately, the land bill is severely faded, and the area is unreadable. Luckily, she knows the areas of the two square plots next to it. Knowing that Emily is good at math, Auntie Agent asks her for help.

Help Emily and Auntie Agent find the area of the new plot.

Hint

Use the formula for the area of a square and the formula for the area of a rectangle.

Solution

The area A of the new plot can be found by using the formula for the area of a rectangle. A=l w Analyzing the diagram, it can be realized that l and w correspond to the lengths of the square plots.

Since the areas of the square plots are known, it is possible to find l and w.

Area of Square Plot Side Length
l ^2=160 l=sqrt(160)
w^2=360 w=sqrt(360)
Finally, the area of the newly acquired plot can be found by substituting these values into the formula for the area of a rectangle. Then, the Product Property of Square Roots can be used.
A=l w
SubstituteII

l= sqrt(160), w= sqrt(360)

A= sqrt(160)* sqrt(360)
ProdSqrt

sqrt(a)*sqrt(b)=sqrt(a* b)

A=sqrt(160* 360)
Multiply

Multiply

A=sqrt(57 600)
CalcRoot

Calculate root

A=240
Therefore, the area of the new plot is 240 square meters. Auntie Agent is ready to wheel and deal!
Example

Simplifying Square Roots

Auntie Agent finds herself bored of the family gathering. She sneaks off to the kitchen wanting to calculate a few math problems from her kid's math textbook! She notices an interesting expression on a graphing calculator.

She notices that the square root of 8 appears to be twice the value of the square root of 2. Auntie Agent, curious to know why, checks her kid's notes and sees the following notes from his class.

The teacher said that the radicand ought to be factored using perfect squares. Then, the Product Property of Square Roots can be used. The teacher suggested to simplify sqrt(18) using this method. Help Auntie Agent rewrite sqrt(18) in terms of sqrt(2). Write the exact value, not an approximation.

Hint

Factor 18 using perfect squares.

Solution
In order to use the Product Property of Square Roots, the radicand should be factored using perfect squares. 18 = 9* 2 Knowing this, the Product Property of Square Roots can be used.
sqrt(18)
SplitIntoFactors

Split into factors

sqrt(9(2))
SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

sqrt(9) * sqrt(2)
CalcRoot

Calculate root

3sqrt(2)
Therefore, sqrt(18) equals 3sqrt(2). Auntie Agent feels relieved to have figured out what the graphing calculator expressed.
Pop Quiz

Simplify the Expression

Use the Product Property of Square Roots to simplify the given square roots.

Simplify Radicand Random
Discussion

Quotient Property of Square Roots

When working with square roots, just like how the product of a square root operates, there is a similar property for quotients.

Let a be a non-negative number and b be a positive number. The square root of the quotient ab equals the quotient of the square roots of a and b.


sqrt(a/b) = sqrt(a)/sqrt(b), for a≥ 0, b> 0

Proof
Let x, y, and z be non-negative numbers such that x=sqrt(a), y=sqrt(b), and z = sqrt(ab). By the definition of a square root, each of these numbers squared is equal to its corresponding radicand. x^2 = a & (I) y^2 = b & (II) z^2 = a/b & (III) Since b is a positive number, y^2 is also positive. Therefore, Equation (I) can be divided by y^2. x^2=& a ⇓ x^2/y^2=& a/y^2 Now, substitute Equations (II) and (III) into this equation.
x^2/y^2 = a/y^2
▼
Substitute values and simplify
Substitute

y^2= b

x^2/y^2 = a/b
Substitute

a/b= z^2

x^2/y^2 = z^2

a^m/b^m=(a/b)^m

(x/y)^2 = z^2
RearrangeEqn

Rearrange equation

z^2=(x/y)^2
Since x, y, and z are non-negative, the final equation implies that z= xy. z^2=(x/y)^2 ⇒ z=x/y The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=x/y ⇔ sqrt(a/b)=sqrt(a)/sqrt(b)
Example

Finding the Hypotenuse of a Right Triangle

Emily roams over to see what her cousins are up to, and one of them is working on some geometry homework. They need to find the hypotenuse of the right triangle shown in the diagram.

Emily's cousin knows that the Pythagorean Theorem can be used to find the hypotenuse c of the triangle. After some algebraic manipulation they managed to isolate c. a^2+b^2=c^2 ⇓ c=sqrt(a^2+b^2) After adding the squares of the legs, they are left with a numeric expression for the hypotenuse of the triangle. They wonder if this can be simplified.

Help Emily and their cousin by simplifying the expression for the hypotenuse of the triangle.

Hint

Use the Quotient Property of Square Roots.

Solution
Emily and their cousin already did most of the work! In order to simplify the given expression the Quotient Property of Square Roots can be used. Then, the square roots of the numerator and denominator can be calculated.
c=sqrt(25/49)
SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

c=sqrt(25)/sqrt(49)
CalcRoot

Calculate root

c=5/7
Therefore, the length of the hypotenuse is 57.
Pop Quiz

Simplify the Square Root

Use the Quotient Property of Square Roots to simplify the given square root.

Simplify Radicand Fraction Random
Discussion

Rationalizing a Denominator

The square root of an irreducible fraction whose denominator is a perfect square will result in a numeric expression with an integer denominator. However, a fraction can have a denominator that is not a perfect square. 3/sqrt(6) If the square root of such a fraction is calculated, the denominator will be an irrational number. There is a way of avoiding irrational numbers in a denominator.

When a fraction has a radical denominator that is an irrational number, it is convenient to rewrite the fraction so that the denominator is an integer. This process is known as rationalization. Consider the following example. 3/sqrt(6) The numeric expression can be rationalized by multiplying the numerator and denominator by the same factor — a factor that removes the radical from the denominator. The fraction is then simplified, if possible.
1
Multiply by the Denominator
expand_more

To remove the radical in the denominator, the numerator and the denominator are both multiplied by the same radical. In this case, multiply by sqrt(6). 3*sqrt(6)/sqrt(6)*sqrt(6)

2
Simplify the Fraction
expand_more
The fraction can now be simplified.
3*sqrt(6)/sqrt(6)*sqrt(6)
MultSqrt

sqrt(a)* sqrt(a)= a

3*sqrt(6)/6
ReduceFrac

a/b=.a /3./.b /3.

sqrt(6)/2
The fraction has now been rationalized because the denominator is an integer number.
Pop Quiz

Rationalizing Denominators

Rationalize the denominator of the given numeric expression.

Rationalize Denominator Random
Discussion

Irrational Conjugates and Rationalization

Some irrational numbers are written as expressions involving rational and irrational terms. As performed previously with perfect squares, these expressions cannot be further simplified into one term and are left written as a sum or difference. Consider the following example. 3+sqrt(2) Given an irrational number in this form, it is possible to find another irrational number by switching the sign of the irrational term.

Concept

Irrational Conjugate

Let a, b, and c be rational numbers, with sqrt(c) irrational. The irrational conjugate of a+bsqrt(c) is obtained by switching the sign of the irrational term. ccc Number & & Conjugate [0.15cm] a+bsqrt(c) & switch sign & a-bsqrt(c) [0.15cm] a-bsqrt(c) & switch sign & a+bsqrt(c)

For example, the conjugate of 2-5sqrt(3) is 2+5sqrt(3).

When the denominator of a numeric expression has a number in this form, it can be rationalized by following a standard procedure.

Method

Rationalizing a Denominator Using Conjugates

When a numeric expression has a denominator that is an irrational radical expression, it is convenient to rewrite the expression so that the denominator is a rational number. This process is known as rationalization. Consider the following example. 2/5+sqrt(3) Since the denominator involves a rational and a square root, the fraction can be rewritten by multiplying both the numerator and denominator by the irrational conjugate of the denominator. The resulting expression can then be simplified.
1
Multiply by the Conjugate of the Denominator
expand_more

To get rid of the radical number in the denominator, the numerator and the denominator are multiplied by its irrational conjugate. In this case, multiply by 5-sqrt(3). 2( 5-sqrt(3))/(5+sqrt(3))( 5-sqrt(3))

2
Simplify the Expression
expand_more
The obtained expression can now be simplified.
2(5-sqrt(3))/(5+sqrt(3))(5-sqrt(3))
ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

2(5-sqrt(3))/5^2-(sqrt(3))^2
CalcPow

Calculate power

2(5-sqrt(3))/25-(sqrt(3))^2
PowSqrt

( sqrt(a) )^2 = a

2(5-sqrt(3))/25-3
SubTerm

Subtract term

2(5-sqrt(3))/22
ReduceFrac

a/b=.a /2./.b /2.

5-sqrt(3)/11
The expression has now been rationalized because the denominator is an integer number.
Example

Simplifying a Ratio that Involves Radicals

Emily now goes over to her cousin Dylan, who looks bored. He says he would rather be painting. She has an idea to cheer him up and shows him the phenomenon of free fall. She walks to the top of the stairs and starts dropping stuff!

Emily then wants to teach Dylan how to simplify a numeric expression which represents the ratio of the free fall time of two different objects. sqrt(1)+sqrt(4)/sqrt(9)+sqrt(2) What is the simplest way of writing this expression?

Hint

Rationalize the denominator using the irrational conjugate.

Solution
Note that 1, 4, and 9 are perfect squares. Therefore, their square roots can be calculated.
sqrt(1)+sqrt(4)/sqrt(9)+sqrt(2)
CalcRoot

Calculate root

1+2/3+sqrt(2)
AddTerms

Add terms

3/3+sqrt(2)
Since the denominator of this expression is an irrational number, it needs to be rationalized in order to be written in its simplest form. This can be done by multiplying the numerator and denominator by the irrational conjugate of the denominator. 3/3 + sqrt(2) * 3 - sqrt(2)/3 - sqrt(2) The expression can be now be simplified.
3/3+sqrt(2) * 3-sqrt(2)/3-sqrt(2)
MultFrac

Multiply fractions

3(3-sqrt(2))/(3+sqrt(2))(3-sqrt(2))
Distr

Distribute 3

9-3sqrt(2)/(3+sqrt(2))(3-sqrt(2))
▼
Simplify denominator
ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

9-3sqrt(2)/3^2-(sqrt(2))^2
CalcPow

Calculate power

9-3sqrt(2)/9-(sqrt(2))^2
PowSqrt

( sqrt(a) )^2 = a

9-3sqrt(2)/9-2
SubTerm

Subtract term

9-3sqrt(2)/7
Discussion

Adding and Subtracting Radicals

Some irrational numbers are written as expressions involving the sum of irrational terms. Below is an example. 3sqrt(2)+5sqrt(2) Given an expression in this form, it is possible — under certain circumstances — to simplify it into one term.

A numeric or algebraic expression that contains two or more radical terms with the same radicand and the same index can be simplified by adding or subtracting the corresponding coefficients.


asqrt(x)± bsqrt(x)=(a± b)sqrt(x)

Here, a,b, and x are real numbers and n is a natural number. If n is even, then x must be greater than or equal to zero.

Proof
If n is an even number, then x is greater than or equal to zero. If n is odd, then x can be any real number. In both cases, sqrt(x) is a real number. This means that the Distributive Property can be used to factor out sqrt(x).
asqrt(x)± bsqrt(x)
FactorOut

Factor out sqrt(x)

(a± b)sqrt(x)

If the radicands are not the same but one or both radical terms can be rewritten to have the same radicand, the expression can be simplified by adding or subtracting radicals. Consider the following example. 3sqrt(8)+2sqrt(2) This expression can be simplified by rewriting sqrt(8) in terms of sqrt(2).

3sqrt(8)+2sqrt(2)
▼
Simplify
SplitIntoFactors

Split into factors

3sqrt(4*2)+2sqrt(2)
SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

3sqrt(4)sqrt(2)+2sqrt(2)
CalcRoot

Calculate root

3*2sqrt(2)+2sqrt(2)
Multiply

Multiply

6sqrt(2)+2sqrt(2)
6sqrt(2)+2sqrt(2)
AddTerms

Add terms

8sqrt(2)
Example

Using the Properties of Square Roots to Simplify an Expression

Emily's cousins Ali and Davontay want to ace next week's math test. An exercise on their study guide asks them to simplify the following numeric expression.

3sqrt(3)+sqrt(12)/sqrt(24)-sqrt(6) Ali says that, since there are no like radicals, the expression cannot be simplified. Davontay suggests the use of the properties of square roots to see if any like radicals appear.

Simplify Radicals Bubble Text
Help Ali and Davontay simplify the given expression!

Hint

Use the Product Property of Square Roots to find like radicals. Simplify by rationalizing the denominator.

Solution
In order to add radicals they need to have the same radicand. The Product Property of Square Roots can be used in the numerator and the denominator to find like radicals.
3sqrt(3)+sqrt(12)/sqrt(24)-sqrt(6)
▼
Simplify numerator
SplitIntoFactors

Split into factors

3sqrt(3)+sqrt(4(3))/sqrt(24)-sqrt(6)
SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

3sqrt(3)+2sqrt(3)/sqrt(24)-sqrt(6)
AddTerms

Add terms

5sqrt(3)/sqrt(24)-sqrt(6)
▼
Simplify denominator
SplitIntoFactors

Split into factors

5sqrt(3)/sqrt(4(6))-sqrt(6)
SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

5sqrt(3)/2sqrt(6)-sqrt(6)
SubTerm

Subtract term

5sqrt(3)/sqrt(6)
Finally, the denominator will be rationalized.
5sqrt(3)/sqrt(6)
▼
Simplify
MovePartNumLeft

a* b/c=a*b/c

5sqrt(3)/sqrt(6)
DivSqrt

sqrt(a)/sqrt(b)=sqrt(a/b)

5sqrt(3/6)
ReduceFrac

a/b=.a /3./.b /3.

5sqrt(1/2)
SqrtQuot

sqrt(a/b)=sqrt(a)/sqrt(b)

51/sqrt(2)
MoveLeftFacToNum

a*b/c= a* b/c

5/sqrt(2)
ExpandFrac

a/b=a * sqrt(2)/b * sqrt(2)

5sqrt(2)/sqrt(2)*sqrt(2)
MultSqrt

sqrt(a)* sqrt(a)= a

5sqrt(2)/2
Closure

Finding the Dimensions of a Golden Rectangle

The challenge presented at the beginning can be solved by using the mathematical tools provided in this lesson. Recall that Dylan is trying to make a canvas that has the dimensions of a golden rectangle.

Dylan Canvas Golden Ratio
In order to do so, Dylan can cut off the excess of his canvas so that the ratio of the dimensions is the golden ratio. l/w=1+sqrt(5)/2 If he cuts the length l of the canvas to be 60 inches. What should be the width w? Round the answer to two decimal places.

Hint

Substitute 60 for l and solve for w.

Solution
The desired width w of the canvas can be found by substituting the known length 60 for l into the equation and solving for w.
l/w=1+sqrt(5)/2
Substitute

l= 60

60/w=1+sqrt(5)/2
▼
Solve for w
MultEqn

LHS * 2=RHS* 2

60/w(2)=1+sqrt(5)
MoveRightFacToNum

a/c* b = a* b/c

120/w=1+sqrt(5)
MultEqn

LHS * w=RHS* w

120=(1+sqrt(5))w
DivEqn

.LHS /(1+sqrt(5)).=.RHS /(1+sqrt(5)).

120/1+sqrt(5)=w
RearrangeEqn

Rearrange equation

w=120/1+sqrt(5)
The right-hand side of the equation can be simplified by rationalizing the denominator using conjugates.
w=120/1+sqrt(5)
▼
Simplify right-hand side
ExpandFrac

a/b=a * (1-sqrt(5))/b * (1-sqrt(5))

w=120(1-sqrt(5))/(1+sqrt(5))(1-sqrt(5))
ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

w=120(1-sqrt(5))/1-(sqrt(5))^2
PowSqrt

( sqrt(a) )^2 = a

w=120(1-sqrt(5))/1-5
SubTerm

Subtract term

w=120(1-sqrt(5))/-4
MoveNegNumToFrac

Put minus sign in front of fraction

w=-120(1-sqrt(5))/4
MovePartNumRight

a* b/c=a/c* b

w=-120/4(1-sqrt(5))
CalcQuot

Calculate quotient

w=-30(1-sqrt(5))
UseCalc

Use a calculator

w=37.082039...
RoundDec

Round to 2 decimal place(s)

w≈37.08


Level 1
Level 2
Level 3
Operating with Square Roots
Exercise 2.1

An appliance used by electrical engineers is capable of measuring the electric current I (in amperes) in terms of the power P (in watts) and the resistance R (in ohms).

T
These three quantities are related by the following formula. I=sqrt(P/R) Find the current I used by an appliance when the power is 120 watts and the resistance is 4 ohms. Round the answer to two decimal places.

We are given that the power P is 120 watts and the resistance R is 4 ohms. Let's substitute these values into the formula!

Therefore, the electric current is about 5.48 amperes.

E
Hint & Answer
S
Solution

Rationalize the following expressions.

1/sqrt(5)
3/sqrt(6)
2/3+sqrt(7)

To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root that will give us a perfect square under the radical.


Let's rationalize this expression by multiplying the numerator and denominator by sqrt(6).


To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.

Binomial Conjugate
a + b a - b
a - b a + b

In this case, the conjugate of the denominator is 3-sqrt(7).


E
Hint & Answer
S
Solution

Tearrik has simplified a rational expression by following the steps shown below. However, his teacher told him that the solution is incorrect.

Calculation steps made by Tearrik
Which of the four steps is wrong?

We will simplify the given expression step by step by using the properties of square roots. This will let us compare the result from each step with Tearrik's solution to identify the mistake. Begin by splitting 75 into at least one perfect square factor.

So far so good. Next, we apply the Product Property of Square Roots.

We note that this step is correct too. We now calculate the square root of 25.

This step is also correct. Since 3 is a prime number and the expression cannot be further simplified, this should be the answer. However, Tearrik went further and made a mistake by multiplying 5 and 3 and taking the square root of the product. Therefore, the mistake is in step 4.

E
Hint & Answer
S
Solution

Operating with Square Roots

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