5.7 Simplify and Use Square Roots

Learning objectives

Learning objectives

• Simplify expressions with square roots

• Estimate square roots

• Approximate square roots

• Simplify variable expressions with square roots

• Use square roots in applications

Simplify Expressions with Square Roots

To start this section, we need to review some important vocabulary and notation.

Remember that when a number $n$ is multiplied by itself, we can write this as ${n}^{2},$ which we read aloud as $“n\,\mathrm{squared."}$ For example, ${8}^{2}$ is read as $\mathrm{"8}\,\mathrm{squared."}$

We call $64$ the square of $8$ because ${8}^{2}=64.$ Similarly, $121$ is the square of $11,$ because ${11}^{2}=121.$

Do you know why we use the word square? If we construct a square with three tiles on each side, the total number of tiles would be nine.

This is why we say that the square of three is nine.

The number $9$ is called a perfect square because it is the square of a whole number.

The chart shows the squares of the counting numbers $1$ through $15.$ You can refer to it to help you identify the perfect squares.

What happens when you square a negative number?

When we multiply two negative numbers, the product is always positive. So, the square of a negative number is always positive.

The chart shows the squares of the negative integers from $-1$ to $-15.$

Did you notice that these squares are the same as the squares of the positive numbers?

Sometimes we will need to look at the relationship between numbers and their squares in reverse. Because ${10}^{2}=100,$ we say $100$ is the square of $10.$ We can also say that $10$ is a square root of $100.$

Notice ${(-10)}^{2}=100$ also, so $-10$ is also a square root of $100.$ Therefore, both $10$ and $-10$ are square roots of $100.$

So, every positive number has two square roots: one positive and one negative.

What if we only want the positive square root of a positive number? The radical sign, $\sqrt{\phantom{0}},$ stands for the positive square root. The positive square root is also called the principal square root.

We can also use the radical sign for the square root of zero. Because ${0}^{2}=0,\sqrt{0}=0.$ Notice that zero has only one square root.

The chart shows the square roots of the first $15$ perfect square numbers.

Example 5.69

Solution

1. Use the definition of a square root. The principal square root of $25$ is the nonnegative number whose square is $25$.
2. Identify the first perfect square. Since $5^2=25$, the principal square root of $25$ is $5$, not the negative number $-5$.
3. Identify the second perfect square. Since $11^2=121$, the principal square root of $121$ is $11$, not the negative number $-11$.
4. Write both simplified values. Therefore, $\sqrt{25}=5$ and $\sqrt{121}=11$, because a radical symbol denotes the nonnegative square root.

Takeaway: The radical symbol names the nonnegative square root, so $\sqrt{25}=5$ and $\sqrt{121}=11$ because those positive numbers square to the radicands.

Every positive number has two square roots and the radical sign indicates the positive one. We write $\sqrt{100}=10.$ If we want to find the negative square root of a number, we place a negative in front of the radical sign. For example, $-\sqrt{100}=-10.$

Example 5.70

Solution

1. Separate the outside sign. In $-\sqrt{9}$ and $-\sqrt{144}$, the negative sign multiplies the entire principal square root rather than the radicand.
2. Evaluate the first radical. Since $3^2=9$, the principal square root $\sqrt{9}$ equals $3$.
3. Evaluate the second radical. Since $12^2=144$, the principal square root $\sqrt{144}$ equals $12$.
4. Apply the negative signs. Multiplying those square roots by $-1$ gives $-\sqrt{9}=-3$ and $-\sqrt{144}=-12$.

Takeaway: A negative sign outside a radical negates the nonnegative square root, giving $-\sqrt{9}=-3$ and $-\sqrt{144}=-12$.

Square Root of a Negative Number

Can we simplify $\sqrt{-25}?$ Is there a number whose square is $-25?$

None of the numbers that we have dealt with so far have a square that is $-25.$ Why? Any positive number squared is positive, and any negative number squared is also positive. In the next chapter we will see that all the numbers we work with are called the real numbers. So we say there is no real number equal to $\sqrt{-25}.$ If we are asked to find the square root of any negative number, we say that the solution is not a real number.

Example 5.71

Solution

1. Inspect the first radicand. The expression $\sqrt{-169}$ asks for a real number whose square is $-169$, but every square of a real number is nonnegative.
2. Classify the first expression. Since no real number has square $-169$, the expression $\sqrt{-169}$ is not a real number.
3. Inspect the second expression. In $-\sqrt{121}$, the radicand $121$ is positive and the negative sign is outside the radical.
4. Evaluate the second expression. Since $11^2=121$, taking the principal square root and then its opposite gives $-\sqrt{121}=-11$.

Takeaway: A negative radicand has no real square root, but a negative sign outside a valid radical gives $-\sqrt{121}=-11$.

Square Roots and the Order of Operations

When using the order of operations to simplify an expression that has square roots, we treat the radical sign as a grouping symbol. We simplify any expressions under the radical sign before performing other operations.

Example 5.72

Solution

1. Separate the radicals in part a. The expression $\sqrt{25}+\sqrt{144}$ contains two independent radical terms with no radical enclosing the addition.
2. Simplify each radical. Evaluate the square roots separately to obtain $5+12$, since $5^2=25$ and $12^2=144$.
3. Finish part a. Add the resulting values to get $5+12=17$.
4. Follow the radical in part b. Evaluate the addition inside the single radical first, giving $\sqrt{25+144}=\sqrt{169}=13$, which differs from $17$.

Takeaway: In $\sqrt{25}+\sqrt{144}$ the separate roots are added to obtain $17$, but in $\sqrt{25+144}$ the radicand is added first to obtain $13$.

Notice the different answers in parts ⓐ and ⓑ of Example 5.72. It is important to follow the order of operations correctly. In ⓐ , we took each square root first and then added them. In ⓑ , we added under the radical sign first and then found the square root.

Estimate Square Roots

So far we have only worked with square roots of perfect squares. The square roots of other numbers are not whole numbers.

We might conclude that the square roots of numbers between $4$ and $9$ will be between $2$ and $3,$ and they will not be whole numbers. Based on the pattern in the table above, we could say that $\sqrt{5}$ is between $2$ and $3.$ Using inequality symbols, we write

Example 5.73

Solution

1. Find nearby perfect squares. The consecutive perfect squares around $60$ are $7^2=49$ and $8^2=64$, so $49<60<64$.
2. Compare the square roots. Because the square-root function preserves order for nonnegative numbers, the corresponding roots satisfy $7<8$.
3. Transfer the bounds to the radical. The position of $60$ between $49$ and $64$ implies that $\sqrt{60}$ lies between $7$ and $8$.
4. Check the estimate. Squaring $7$ and $8$ recovers the original perfect-square bounds, confirming that $7<\sqrt{60}<8$.

Takeaway: Since $49<60<64$, the square root lies between $7$ and $8$, so $7<\sqrt{60}<8$.

Approximate Square Roots with a Calculator

There are mathematical methods to approximate square roots, but it is much more convenient to use a calculator to find square roots. Find the $\sqrt{\phantom{0}}$ or $\sqrt{x}$ key on your calculator. You will need to use this key to approximate square roots. When you use your calculator to find the square root of a number that is not a perfect square, the answer that you see is not the exact number. It is an approximation, to the number of digits shown on your calculator’s display. The symbol for an approximation is $\approx$ and it is read approximately.

Suppose your calculator has a $\mathrm{10-digit}$ display. Using it to find the square root of $5$ will give $2.236067977.$ This is the approximate square root of $5.$ When we report the answer, we should use the “approximately equal to” sign instead of an equal sign.

You will seldom use this many digits for applications in algebra. So, if you wanted to round $\sqrt{5}$ to two decimal places, you would write

How do we know these values are approximations and not the exact values? Look at what happens when we square them.

The squares are close, but not exactly equal, to $5.$

Example 5.74

Solution

1. Enter the radicand. Use the square-root key on a calculator with $17$ as the input, taking care to evaluate $\sqrt{17}$ rather than $17^2$.
2. Read the displayed value. The calculator gives approximately $4.123105626$, which is a decimal approximation of the radical.
3. Locate the rounding place. Two decimal places means keeping the hundredths digit and using the thousandths digit to determine the rounding action.
4. Round and state the result. The thousandths digit is $3$, so the value rounds down to $4.12$ and $\sqrt{17}\approx4.12$ to two decimal places.

Takeaway: A calculator value of $4.123105626$ rounds at the hundredths place to give $\sqrt{17}\approx4.12$ to two decimal places.

Simplify Variable Expressions with Square Roots

Expressions with square root that we have looked at so far have not had any variables. What happens when we have to find a square root of a variable expression?

Consider $\sqrt{9{x}^{2}},$ where $x\ge{}0.$ Can you think of an expression whose square is $9{x}^{2}?$

When we use a variable in a square root expression, for our work, we will assume that the variable represents a non-negative number. In every example and exercise that follows, each variable in a square root expression is greater than or equal to zero.

Example 5.75

Solution

1. Inspect the radicand. The expression $\sqrt{x^2}$ contains the perfect square $x^2$, so look for a nonnegative quantity whose square equals $x^2$.
2. Use the variable assumption. The problem states that $x$ is nonnegative, so $x$ itself is the nonnegative quantity needed for the principal square root.
3. Apply the square-root rule. Because $(x)^2=x^2$ and $x$ is nonnegative, the principal square root of $x^2$ is $x$.
4. Check the simplified expression. Squaring the result $x$ returns the original radicand $x^2$, and the nonnegative assumption confirms the correct sign.

Takeaway: With $x$ assumed nonnegative, the perfect square inside the radical gives $\sqrt{x^2}=x$.

Example 5.76

Solution

1. Find the perfect-square coefficient. The number $16$ is $4^2$, while the remaining factor $x^2$ is already a squared variable.
2. Rewrite the radicand. Combine these factors as a product square: $16x^2=(4x)^2$.
3. Take the principal square root. The radical therefore gives the nonnegative square root of $(4x)^2$, which is $4x$ under the problem's variable assumptions.
4. Verify the result. Squaring $4x$ gives $16x^2$, confirming that $\sqrt{16x^2}=4x$ and that no sign was lost.

Takeaway: Because $16x^2=(4x)^2$, the principal square root of $\sqrt{16x^2}$ is $4x$.

Example 5.77

Solution

1. Keep the negative outside. The expression $-\sqrt{81y^2}$ has a negative coefficient before the radical, so that sign applies to the completed square root.
2. Find the perfect-square coefficient. The number $81$ is $9^2$, and the other factor $y^2$ is already a squared variable.
3. Rewrite the radicand. Combine the factors as $81y^2=(9y)^2$ so the expression becomes $-\sqrt{(9y)^2}$.
4. Evaluate and apply the sign. The principal square root is $9y$ under the given variable assumptions, and the outside negative makes the final result $-9y$.

Takeaway: Simplifying the radical in $-\sqrt{81y^2}$ gives $9y$ first, and the outside negative then produces $-9y$.

Example 5.78

Solution

1. Identify every squared factor. The radicand $36x^2y^2$ contains the perfect-square coefficient $36$ together with the squared variables $x^2$ and $y^2$.
2. Choose the matching square roots. The positive square root of $36$ is $6$, while the nonnegative square roots associated with $x^2$ and $y^2$ are $x$ and $y$.
3. Form the complete square. Multiplying those factors gives $36x^2y^2=(6xy)^2$, which reveals the radicand as one perfect square.
4. Take the principal square root. The radical simplifies to $6xy$, and squaring this result returns the original radicand as a check.

Takeaway: The perfect-square factors in $36x^2y^2$ combine as $(6xy)^2$, so its principal square root is $6xy$.

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter several applications of square roots. Once again, if we use our strategy for applications, it will give us a plan for finding the answer!

We have solved applications with area before. If we were given the length of the sides of a square, we could find its area by squaring the length of its sides. Now we can find the length of the sides of a square if we are given the area, by finding the square root of the area.

If the area of the square is $A$ square units, the length of a side is $\sqrt{A}$ units. See Table 5.7.

Table

5.7

Example 5.79

Solution

1. Connect area and side length. The area of a square equals its side length squared, so a side length $s$ satisfies $s^2=200$.
2. Solve for the side length. Taking the principal square root of both sides gives $s=\sqrt{200}$ in feet.
3. Evaluate the radical. A calculator gives $\sqrt{200}\approx14.142135$, which is the decimal side length before rounding.
4. Round to the requested precision. To the nearest tenth of a foot, the patio's side can be $14.1$ feet long.

Takeaway: The side of the $200$ square foot patio is $\sqrt{200}\approx14.1$ feet when rounded to the nearest tenth.

Square Roots and Gravity

Another application of square roots involves gravity. On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it will take to reach the ground is found by evaluating the expression $\frac{\sqrt{h}}{4}.$ For example, if an object is dropped from a height of $64$ feet, we can find the time it takes to reach the ground by evaluating $\frac{\sqrt{64}}{4}.$

It would take $2$ seconds for an object dropped from a height of $64$ feet to reach the ground.

Example 5.80

Solution

1. Use the fall-time model. The time in seconds is modeled by $\frac{\sqrt{h}}{4}$, where $h$ is the initial height in feet.
2. Substitute the height. Replace $h$ with $400$ to obtain the expression $\frac{\sqrt{400}}{4}$ for the time to reach the river.
3. Simplify the radical. Since $20^2=400$, the principal square root $\sqrt{400}$ equals $20$.
4. Divide and interpret. The expression becomes $\frac{20}{4}=5$, so the sunglasses take $5$ seconds to reach the river.

Takeaway: Substituting $400$ feet into the fall-time model gives $\frac{\sqrt{400}}{4}=\frac{20}{4}=5$ seconds.

Square Roots and Accident Investigations

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes. According to some formulas, if the length of the skid marks is $d$ feet, then the speed of the car can be found by evaluating $\sqrt{24d}.$

Example 5.81

Solution

1. Use the skid-distance model. The car's speed in miles per hour is modeled by $\sqrt{24d}$, where $d$ is the skid distance in feet.
2. Substitute the measured distance. Replace $d$ with $190$ to obtain $\sqrt{24\cdot190}$ as the speed before the brakes were applied.
3. Multiply and evaluate. The product is $4560$, and a calculator gives $\sqrt{4560}\approx67.527772$ miles per hour.
4. Round to the nearest tenth. The speed rounds to $67.5$ miles per hour, preserving the model's unit and the requested precision.

Takeaway: The model $\sqrt{24\cdot190}$ evaluates to about $67.5$ miles per hour for the $190$ foot skid distance.

Section 5.7 Exercises

Section 5.7 Exercises

Simplify Expressions with Square Roots

In the following exercises, simplify.

Estimate Square Roots

In the following exercises, estimate each square root between two consecutive whole numbers.

Approximate Square Roots with a Calculator

In the following exercises, use a calculator to approximate each square root and round to two decimal places.

Simplify Variable Expressions with Square Roots

In the following exercises, simplify. (Assume all variables are greater than or equal to zero.)

Use Square Roots in Applications

In the following exercises, solve. Round to one decimal place.

Everyday Math

Writing Exercises

Self Check

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

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

1. Simplify: ⓐ $\,\sqrt{36}\,$ⓑ $\,\sqrt{169}.$

2. Simplify: ⓐ $\,\sqrt{16}\,$ⓑ $\,\sqrt{196}.$

3. Simplify: ⓐ $\,-\sqrt{4}\,$ⓑ $\,-\sqrt{225}.$

1. Simplify: ⓐ $\,-\sqrt{81}\,$ⓑ $\,-\sqrt{64}.$

2. Simplify: ⓐ $\,\sqrt{-196}\,$ⓑ $\,-\sqrt{81}.$

3. Simplify: ⓐ $\,\sqrt{-49}\,$ⓑ $\,-\sqrt{121}.$

1. Simplify: ⓐ $\,\sqrt{9}+\sqrt{16}\,$ⓑ $\,\sqrt{9+16}.$

2. Simplify: ⓐ $\,\sqrt{64+225}\,$ⓑ $\,\sqrt{64}+\sqrt{225}.$

3. Estimate $\sqrt{38}$ between two consecutive whole numbers.

1. Estimate $\sqrt{84}$ between two consecutive whole numbers.

2. Round $\sqrt{11}$ to two decimal places.

3. Round $\sqrt{13}$ to two decimal places.

1. Simplify: $\sqrt{{y}^{2}}.$

2. Simplify: $\sqrt{{m}^{2}}.$

3. Simplify: $\sqrt{64{x}^{2}}.$

1. Simplify: $\sqrt{169{y}^{2}}.$

2. Simplify: $-\sqrt{121{y}^{2}}.$

3. Simplify: $-\sqrt{100{p}^{2}}.$

1. Simplify: $\sqrt{100{a}^{2}{b}^{2}}.$

2. Simplify: $\sqrt{225{m}^{2}{n}^{2}}.$

3. Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of $370$ square feet. To the nearest tenth of a foot, how long can a side of her square lawn be?

1. Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of $2704$ square centimeters. How long can a side of his mosaic be?

2. A helicopter drops a rescue package from a height of $1296$ feet. How many seconds does it take for the package to reach the ground?

3. A window washer drops a squeegee from a platform $196$ feet above the sidewalk. How many seconds does it take for the squeegee to reach the sidewalk?

1. An accident investigator measured the skid marks of a car and found their length was $76$ feet. To the nearest tenth, what was the speed of the car before the brakes were applied?

2. The skid marks of a vehicle involved in an accident were $122$ feet long. To the nearest tenth, how fast had the vehicle been going before the brakes were applied?

3. $-\sqrt{4}$

1. $-\sqrt{1}$

2. $\sqrt{-121}$

3. $\sqrt{-9}$

1. $\sqrt{70}$

2. $\sqrt{200}$

3. $\sqrt{{y}^{2}}$

1. $\sqrt{49{x}^{2}}$

2. $-\sqrt{64{a}^{2}}$

3. $\sqrt{144{x}^{2}{y}^{2}}$

1. Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of $75$ square feet. How long can a side of his garden be?

2. Gravity An airplane dropped a flare from a height of $1,024$ feet above a lake. How many seconds did it take for the flare to reach the water?

3. Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, $4,000$ feet above the Colorado River. How many seconds did it take for the hammer to reach the river?

1. Accident investigation The skid marks from a car involved in an accident measured $216$ feet. What was the speed of the car before the brakes were applied?

2. Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was $117$ feet. What was the speed of the vehicle before the brakes were applied?

3. Decorating Morris wants to have a square mosaic inlaid in his new patio. His budget allows for $2,025$ tiles. Each tile is square with an area of one square inch. How long can a side of the mosaic be?

1. What is the difference between ${9}^{2}$ and $\sqrt{9}?$