# 1.3: Square and cube roots of real numbers (2023)

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learning objective

• Calculates exact and approximate values ​​for the square root of a real number.
• Calculates exact and approximate values ​​for the cube root of a real number.
• Simplifies square and cube roots of real numbers.
• Apply the Pythagorean theorem.

## Definition of square root and cube root

ONEsquare root74An arithmetic is a number that, when multiplied by itself, produces the original number. For example, $$4$$ is the square root of $$16$$, because $$4^{2}=16$$. Since $$(−4)^{2}=16$$, we can say that $$−4$$ is also the square root of $$16$$. Every positive real number has two square roots, one positive and one negative. That's what we useroot75 $$√$$ meansPrincipal (non-negative) square root76There is a minus sign in front of the root symbol $$−√$$ , which means a negative square root.

$$\sqrt { 16 } = 4 \color{sky blue}{\:square\:root\:of\:16}$$

$$- \sqrt { 16 } = - 4 \color{blue {minus\:square\:root\:of\:16}$$

Zero is the only real number that has exactly one square root.

$$\sqrt{0} = 0$$

ifRoot number77, the number inside the root symbol is nonzero and can be factored into the square of another nonzero number, then the square root of that number is obvious. In this case we have the following properties:

$$\sqrt { a ^ { 2 } } = a , \text { if } a \geq 0$$

It should be noted that $$a$$ must be non-negative. Note $$\sqrt { ( - 3 ) ^ { 2 } } \neq - 3$$ because the root represents the principal square root. Instead,

$$\sqrt { ( - 3 ) ^ { 2 } } = \sqrt { 9 } = 3$$

This distinction will be carefully considered later in this lesson.

Example $$\PageIndex{1}$$:

Find the square root:

1. $$\sqrt { 121 }$$
2. $$\sqrt {0,25 }$$
3. $$\sqrt { \frac { 4 } { 9 } }$$

solution

1. $$\sqrt { 121 } = \sqrt { 11 ^ { 2 } } = 11$$
2. $$\sqrt { 0.25 } = \sqrt { 0.5 ^ { 2 } } = 0.5$$
3. $$\sqrt { \frac { 4 } { 9 } } = \sqrt { \left( \frac { 2 } { 3 } \right) ^ { 2 } } = \frac { 2 } { 3 }$$

Example $$\PageIndex{2}$$:

Find the negative square root:

1. $$−\sqrt{64}$$
2. $$−\sqrt{1}$$

solution

1. $$- \sqrt { 64 } = - \sqrt { 8 ^ { 2 } } = - 8$$
2. $$- \sqrt { 1 } = - \sqrt { 1 ^ { 2 } } = - 1$$

The sliced ​​number may not always be a perfect square. If a positive integer is not a perfect square, then its square root will be irrational. Given $$\sqrt{5}$$, we can obtain an approximation by restricting it using perfect squares $$4$$ and $$9$$ as follows:

$$\begin{array} { c } { \sqrt { 4 } < \sqrt { 5 } < \sqrt { 9 } } \\ { 2 < \sqrt { 5 } < 3 } \end{array}$$

From this we conclude that $$\sqrt{5}$$ is between $$2$$ and $$3$$. Use the square root $$√$$ button on most calculators to get a better approximation of this number.

$$\sqrt { 5 } \ about 2.236 \mathrm { because } 2.236 \wedge 2 \about 5$$

Next, consider the square root of a negative number. To determine the square root of $$−9$$, you need to find a number whose square results in $$−9$$,

$$\sqrt { - 9 } = \color{Cerulean}{?}$$ $$\text { 或 } ( \color{Cerulean}{?}$$$$)^ { 2 } = - 9$$

However, the square of any real number always gives a positive number,

$$( 3 ) ^ { 2 } = 9 \text { and } ( - 3 ) ^ { 2 } = 9$$

The square root of negative numbers is currently undefined. Try calculating $$\sqrt{-9}$$ on a calculator, what does it say? Now, we will declare that $$\sqrt{−9}$$ is not a real number. The square root of negative numbers will be defined later in the lesson.

ONEcube root78 A number is a number that is tripled by itself to get the original number. Furthermore, we use the notation $$\sqrt [ 3 ] { }$$ to represent the cube root, where $$3$$ is calledindex79. For example,

$$\sqrt [ 3 ] { 8 } = 2 , \text { because } 2 ^ { 3 } = 8$$

The product of three equal factors will be positive if the factors are positive and negative if the factors are negative. Therefore, every real number has exactly one real cube root. Therefore, the technical details related to roots do not apply. For example,

$$\sqrt [ 3 ] { - 8 } = - 2 , \text { because } ( - 2 ) ^ { 3 } = - 8$$

In general, given any real number $$a$$, we have the following properties:

$$\sqrt [ 3 ] { a ^ { 3 } } = a$$

When simplifying a cube root, look for factors of a perfect cube.

Example $$\PageIndex{3}$$

Find the cube root:

1. $$\sqrt [ 3 ] { 125 }$$
2. $$\sqrt [ 3 ] {0 }$$
3. $$\sqrt [ 3 ] { \frac { 8 } { 27 } }$$

solution

1. $$\sqrt [ 3 ] { 125 } = \sqrt [ 3 ] { 5 ^ { 3 } } = 5$$
2. $$\sqrt [ 3 ] { 0 } = \sqrt [ 3 ] { 0 ^ { 3 } } = 0$$
3. $$\sqrt [ 3 ] { \frac { 8 } { 27 } } = \sqrt [ 3 ] { \left( \frac { 2 } { 3 } \right) ^ { 3 } } = \frac { 2 } { 3}$$

Example $$\PageIndex{4}$$

Find the cube root:

1. $$\sqrt [ 3 ] { - 27 }$$
2. $$\sqrt [ 3 ] { - 1 }$$

solution

1. $$\sqrt [ 3 ] { - 27 } = \sqrt [ 3 ] { ( - 3 ) ^ { 3 } } = - 3$$
2. $$\sqrt [ 3 ] { - 1 } = \sqrt [ 3 ] { ( - 1 ) ^ { 3 } } = - 1$$

The cut number may not be a perfect cube. If this is the case, then its cube root will be an irrational number. For example, $$\sqrt [ 3 ] { 2 }$$ is an irrational number that can be approximated using the $$\sqrt [ x] { }$$ root button on most calculators. Depending on the calculator, we usually enter the exponent and then the dividend before pressing the button, like so:

$$3\:\:\: \sqrt [x] {y}\:\:\: 2\:\:\:=$$

Therefore, we have

$$\sqrt [ 3 ] { 2 } \ about 1.260 , \text { for } 1.260 \wedge 3 \ about 2$$

Later in this lesson, we will expand on these ideas using any integer as an index. It should be noted that the index of the square root is $$2$$, so the following are equivalent:

$$\sqrt [ 2 ] { a } = \sqrt { a }$$

## Simplify the square and cube roots

The sliced ​​number is not always a perfect square. If not, we use the following two properties to simplify the expression. Given real numbers $$\sqrt [ n ] { A }$$ and $$\sqrt [ n ] { B }$$, where $$B ≠ 0$$,

• radical product rule:80$\sqrt [ n ] { A \cdot B } = \sqrt [ n ] { A } \cdot \sqrt [ n ] { B }$
• root quotient rule:81$\sqrt [ n ] { \frac { A } { B } } = \frac { \sqrt [ n ] { A } } { \sqrt [ n ] { B } }$

ONEsimplified roots82 is one of those numbers for which the number contains no coefficients that can be written as exponents of perfect powers. Given a square root, the idea is to determine the largest square factor of the number and then apply the properties shown above. For example, to simplify $$\sqrt{12}$$, note that $$12$$ is not a perfect square. However, $$12$$ has a perfect square factor, $$12 = 4 ⋅ 3$$. Apply this property as follows:

\begin{align*} \sqrt { 12 } &= \sqrt { 4 \cdot 3 } \quad\color{Cerulean}{apply\: product\: rules\: to roots. } \ \[4pt] &= \sqrt { 4 } \cdot \sqrt { 3 } \quad\color{sky blue} {simplified} \\[4pt] &= 2 \cdot \sqrt { 3 } \end{ align *}\ ] The number $$2 \sqrt{3}$$ is a simplified irrational number. You are often asked to find an approximate answer, rounded to a certain decimal place. In this case, use a calculator to find decimal approximations using the original problem or a simplified equivalent. $$\sqrt { 12 } = 2 \sqrt { 3 } \approx 3.46$$ As a check, calculate $$\sqrt{12}$$ and $$2\sqrt{3}$$ on a calculator and verify that the results are approximately $$3.46$$. Example $$\PageIndex{5}$$ Simplified: $$\sqrt{135}$$. solution First find the largest perfect square factor $$135$$. \begin{alignment} 135 & = 3 ^ { 3 } \cdot 5 \\ & = 3 ^ { 2 } \cdot 3 \cdot 5 \\ & = 9 \cdot 15 \end{alignment} So, \[ \begin{align*} \sqrt { 135 } &= \sqrt { 9 \cdot 15 } \quad\color{Cerulean}{apply\: product\: rule\: to roots. } \ \[4pt] &= \sqrt { 9 } \cdot \sqrt { 15 } \quad\color{sky blue}{simplified. } \\[4pt] &= 3 \cdot \sqrt { 15 }\end{match*}

$$3\sqrt{15}$$

Example $$\PageIndex{6}$$

Simplified: $$\sqrt { \frac { 108 } { 169 } }$$.

solution

First we find the prime factorization of $$108$$ and $$169$$. This will allow us to easily determine the largest perfect square coefficient.

\begin{match*} 108 & = 2 ^ { 2 } \cdot 3 ^ { 3 } = 2 ^ { 2 } \cdot 3 ^ { 2 } \cdot 3 \\ 169 & = 13 ^ { 2 } \ end {alignment*}

So,

\begin{align*} \sqrt { \frac { 108 } { 169 } } &= \sqrt { \frac { 2 ^ { 2 } \cdot 3 ^ { 2 } \cdot 3 } { 13 ^ { 2 } } }\color{Cerulean}{Apply the product and quotient rule of roots. } \\[4pt] &= \frac { \sqrt { 2 ^ { 2 } } \cdot \sqrt { 3 ^ { 2 } } \cdot \sqrt { 3 } } { \sqrt { 13 ^ { 2 } } \ color{sky blue}simplifies. } \\[4pt] &= \frac { 2 \cdot 3 \cdot \sqrt { 3 } } { 13 } \\ & = \frac { 6 \sqrt { 3 } } { 13 } \end{align*}\ ] answer $$\frac { 6 \sqrt { 3 } } { 13 }$$ Example $$\PageIndex{7}$$ Simplify$$−5\sqrt{162}$$. solution \[ \begin{align*} - 5 \sqrt { 162 } &= - 5 \cdot \sqrt { 81 \cdot 2 } \\[4pt] &= - 5 \cdot \color{天蓝色}{\sqrt { 81 } \cdot \sqrt { 2 }} \\[4pt] &= - 5 \cdot \color{Cerulean}{9 \cdot \sqrt { 2 }} \\[4pt] & = - 45 \cdot \sqrt { 2 } \\[4pt] & = - 45 \sqrt { 2 } \end{align*}

$$−45\sqrt{2}$$

Exercise $$\PageIndex{1}$$

Simplify $$4\sqrt{150}$$

$$20\sqrt{6}$$

A cube root is simplified if it contains no factors that can be written as a perfect cube. The idea is to determine the largest cube factor of the number being measured and then apply the product or quotient rule. For example, to simplify $$\sqrt [ 3 ] { 80 }$$, note that $$80$$ is not a perfect cube. However, $$80 = 8 ⋅ 10$$ we can write,

\begin{align*} \sqrt [ 3 ] { 80 } &= \sqrt [ 3 ] { 8 \cdot 10 }\color{Cerulean}{Application\:Product\:Rules\:Used\:Radials . } \\5pt] &= \sqrt [ 3 ] { 8 } \cdot \sqrt [ 3 ] { 10 }\color{sky blue}{simplified. } \\[4pt] &= 2 \cdot \sqrt [ 3 ] { 10 } \end{match*}

Example $$\PageIndex{8}$$:

Simplify $$\sqrt [ 3 ] { 162 }$$

solution

First find the largest perfect cubic factor $$162$$.

\begin{alignment} 162 & = 3 ^ { 4 } \cdot 2 \\ & = 3 ^ { 3 } \cdot 3 \cdot 2 \\ & = 27 \cdot 6 \end{alignment}

So,

$$\sqrt [ 3 ] { 162 } = \sqrt [ 3 ] { 27 \cdot 6 }\color{Cerulean}{apply\: product\: rules\: to roots.}$$

$$= \sqrt [ 3 ] { 27 } \cdot \sqrt [ 3 ] { 6 }\color{sky blue}{simplified.}$$

$$= 3 \cdot \sqrt [ 3 ] { 6 }$$

$$3 \sqrt [ 3 ] { 6 }$$

Example $$\PageIndex{9}$$:

Simplified: $$\sqrt [ 3 ] { - \frac { 16 } { 343 } }$$.

solution

$$\begin{对齐} \sqrt [ 3 ] { - \frac { 16 } { 343 } } & = \frac { \sqrt [ 3 ] { - 1 \cdot 8 \cdot 2 } } { \sqrt [ 3 ] { 7 ^ { 3 } } } \\ & = \frac { \sqrt [ 3 ] { - 1 } \cdot \sqrt [ 3 ] { 8 } \cdot \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] ] { 7 ^ { 3 } } \\ & = \frac { - 1 \cdot 2 \cdot \sqrt [ 3 ] { 2 } } { 7 } \\ & = \frac { - 2 \sqrt [ 3 ] { 2 } } { 7 } \end{对齐}$$

$$\frac { - 2 \sqrt [ 3 ] { 2 } } { 7 }$$

Exercise $$\PageIndex{2}$$

Simplify $$- 2 \sqrt [ 3 ] { - 256 }$$.

$$8 \sqrt [ 3 ] { 4 }$$

Consider the following two calculations,

$$\begin{array} { l } { \sqrt { 81 } = \sqrt { 9 ^ { 2 } } = 9 } \\ { \sqrt { 81 } = \sqrt { 9 ^ { 2 } } = ( \ sqrt { 9 } ) ^ { 2 } = ( 3 ) ^ { 2 } = 9 } \end{array}$$

Note that it does not matter whether we apply the exponent or the square root first. This is true for any positive real number. We have the following,

$$\sqrt{a^{2}}=(\sqrt{a})^{2}=a,\text{if}a\geq0$$

Example $$\PageIndex{10}$$:

Simplified: $$(\sqrt { 10 } ) ^ { 2 }$$.

solution

If $$a$$ is non-negative, the fact that $$( \sqrt { a } ) ^ { 2 } = a$$ holds.

$$( \sqrt { 10 } ) ^ { 2 } = 10$$

## Pythagorean theorem

ONEright triangle83 is a triangle with one angle $$90°$$. The side opposite the right angle is the longer side and is calledhypotenuse84, the other two sides are calledfoot85. Many real-world applications involve this geometry. ThisPythagorean theorem86 Point out that given any right triangle whose sides are $$a$$ and $$b$$ units, the square of the measure of the hypotenuse$$c$$ is equal to the sum of the squares of the measures of the sides , $$a ^{2} + b^{2} = c^{2}$$. In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its sides.

Example $$\PageIndex{11}$$:

Calculates the diagonal of a square with sides of length $$5$$ units.

solution

The diagonals of the square form an isosceles right triangle with two equal sides each $$5$$ units long.

We can use the Pythagorean theorem to determine the length of the hypotenuse.

\αρχή{στοίχιση} c & = \sqrt { a ^ { 2 } + b ^ { 2 } } \\ & = \sqrt { 5 ^ { 2 } + 5 ^ { 2 } } \\ & = \sqrt { 25 + 25 } \\ & = \sqrt { 50 } \\ & = \sqrt { 25 \cdot 2 } \\ & = \sqrt { 25 } \cdot \sqrt { 2 } \\ & = 5 \cdot \ sqrt { 2 } \end{alignment}

answer: $$5 \sqrt { 2 }$$ units

The Pythagorean theorem actually states that a side length satisfying the property $$a^{2} + b^{2} = c^{2}$$ is a necessary and sufficient condition for a right triangle. In other words, if we can prove that the sum of the squares of the side lengths of a triangle is equal to the square of the hypotenuse, then it must be a right triangle.

Example $$\PageIndex{12}$$:

Determine whether a triangle with sides $$a = 1$$ cm and $$b = 2$$ cm and hypotenuse $$b = \sqrt{5}$$ cm is a right triangle.

solution

The Pythagorean theorem guarantees that a triangle is right angled if the side satisfies the condition $$a^{2} + b^{2} = c^{2}$$.

$$\begin{Alignment} a ^ { 2 } + b ^ { 2 } & = c ^ { 2 } \\ ( 1 ) ^ { 2 } + ( 2 ) ^ { 2 } & = ( \sqrt { 5 } ) ^ { 2 } \\ 1 + 4 & = 5 \\ 5 & = 5 \color{olive green}{✓} \end{match}$$

answer: Yes, the triangle described is right-angled.

## main point

• The square root of a number is the number that is squared to get the original number. The principal square root of a positive real number is the positive square root. The square root of negative numbers is currently undefined.
• When simplifying the square root of a number, find the perfect square factor of the number being numbered. Apply product or quotient rules to the roots and then simplify.
• The cube root of a number is the number cubed to get the original number. Every real number has exactly one real cube root.
• When simplifying the cube root, look for the perfect cube factor of the number. Apply product or quotient rules to the roots and then simplify.
• The Pythagorean theorem gives us the necessary and sufficient conditions for right triangles: $$a^{2} + b^{2} = c^{2}$$ if and only if $$a, b$$ and\ (c \) represents the side length of a right triangle.

Exercise $$\PageIndex{3}$$

1. $$\sqrt{81}$$
2. $$\sqrt{49}$$
3. $$-\sqrt{16}$$
4. $$−\sqrt{100}$$
5. $$\sqrt { \frac { 25 } { 16 } }$$
6. $$\sqrt { \frac { 9 } { 64 } }$$
7. $$\sqrt { \frac { 1 } { 4 } }$$
8. $$\sqrt { \frac { 1 } { 100 } }$$
9. $$\sqrt{-1}$$
10. $$\sqrt{-25}$$
11. $$\sqrt{036}$$
12. $$\sqrt{1.21}$$
13. $$\sqrt{(-5)^{2}}$$
14. $$\sqrt{(-6)^{2}}$$
15. $$2\sqrt{64}$$
16. $$3\sqrt{36}$$
17. $$-10\sqrt{4}$$
18. $$-8\sqrt{25}$$
19. $$\sqrt [ 3 ] { 64 }$$
20. $$\sqrt [ 3 ] { 125 }$$
21. $$\sqrt [ 3 ] { -27 }$$
22. $$\sqrt [ 3 ] { -1 }$$
23. $$\sqrt [ 3 ] {0 }$$
24. $$\sqrt [ 3 ] {0,008 }$$
25. $$\sqrt [ 3 ] {0,064 }$$
26. $$-\sqrt [ 3 ] { -8 }$$
27. $$-\sqrt [ 3 ] { 1000 }$$
28. $$\sqrt [ 3 ] { ( - 8 ) ^ { 3 } }$$
29. $$\sqrt [ 3 ] { ( - 15 ) ^ { 3 } }$$
30. $$\sqrt [ 3 ] { \frac { 1 } { 216 } }$$
31. $$\sqrt [ 3 ] { \frac { 27 } { 64 } }$$
32. $$\sqrt [ 3 ] { -\frac { 1 } { 8 } }$$
33. $$\sqrt [ 3 ] { -\frac { 1 } { 27 } }$$
34. $$5 \sqrt [ 3 ] { 343 }$$
35. $$4 \sqrt [ 3 ] { 512 }$$
36. $$- 10 \sqrt [ 3 ] { 8 }$$
37. $$- 6 \sqrt [ 3 ] { - 64 }$$
38. $$8 \sqrt [ 3 ] { - 8 }$$

1.$$9$$

3.$$−4$$

5.$$\frac{5}{4}$$

7.$$\frac{1}{2}$$

9. Not a real number.

11. $$0,6$$

13.$$5$$

15.$$16$$

17.$$−20$$

19.$$4$$

21.$$−3$$

23.$$0$$

25. $$0,4$$

27.$$−10$$

29.$$−15$$

31.$$\frac{3}{4}$$

33. $$−\frac{1}{3}$$

35.$$32$$

37.$$24$$

Exercise $$\PageIndex{4}$$

Use a calculator to round to the nearest centimeter.

1. $$\sqrt{3}$$
2. $$\sqrt{10}$$
3. $$\sqrt{19}$$
4. $$\sqrt{7}$$
5. $$3\sqrt{5}$$
6. $$-2\sqrt{3}$$
7. $$\sqrt [ 3 ] { 3 }$$
8. $$\sqrt [ 3 ] { 6 }$$
9. $$\sqrt [ 3 ] { 28 }$$
10. $$\sqrt [ 3 ] { 9 }$$
11. $$4\sqrt [ 3 ] { 10 }$$
12. $$-3\sqrt [ 3 ] { 12 }$$
13. Specifies the set consisting of the squares of the first twelve positive integers.
14. Determine the set consisting of the cubes of the first twelve positive integers.

1.$$1,73$$

3.$$4.36$$

5.$$6,71$$

7. $$1,44$$

9.$$3.04$$

11.$$8,62$$

13. $$\{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144\}$$

Exercise $$\PageIndex{5}$$

simplify.

1. $$\sqrt{18}$$
2. $$\sqrt{50}$$
3. $$\sqrt{24}$$
4. $$\sqrt{40}$$
5. $$\sqrt { \frac { 50 } { 81 } }$$
6. $$\sqrt { \frac { 54 } { 25 } }$$
7. $$4 \sqrt { 72 }$$
8. $$3 \sqrt { 27 }$$
9. $$-5 \sqrt { 80 }$$
10. $$-6 \sqrt { 128 }$$
11. $$3 \sqrt { -40 }$$
12. $$5 \sqrt { -160 }$$
13. $$\sqrt [ 3 ] { 16 }$$
14. $$\sqrt [ 3 ] { 54 }$$
15. $$\sqrt [ 3 ] { 81 }$$
16. $$\sqrt [ 3 ] { 24 }$$
17. $$\sqrt [ 3 ] { \frac { 48 } { 125 } }$$
18. $$\sqrt [ 3 ] { \frac { 135 } { 64 } }$$
19. $$7 \sqrt [ 3 ] { 500 }$$
20. $$25 \sqrt [ 3 ] { 686 }$$
21. $$- 2 \sqrt [ 3 ] { - 162 }$$
22. $$5 \sqrt [ 3 ] { - 96 }$$
23. $$( \sqrt { 64 } ) ^ { 2 }$$
24. $$( \sqrt { 25 } ) ^ { 2 }$$
25. $$( \sqrt { 2 } ) ^ { 2 }$$
26. $$( \sqrt { 6 } ) ^ { 2 }$$

1. $$3\sqrt{2}$$

3. $$2\sqrt{6}$$

5. $$\frac { 5 \sqrt { 2 } } { 9 }$$

7. $$24\sqrt{2}$$

9. $$-20\sqrt{5}$$

11. Not a real number.

13. $$2 \sqrt [ 3 ] { 2 }$$

15. $$3 \sqrt [ 3 ] { 3 }$$

17. $$\frac { 2 \sqrt [ 3 ] { 6 } } { 5 }$$

19. $$35 \sqrt [ 3 ] { 4 }$$

21. $$6 \sqrt [ 3 ] { 6 }$$

23. 64

25. 2

Exercise $$\PageIndex{6}$$

1. If the sides of a right triangle measure $$3$$ units and $$4$$ units respectively, find the length of the hypotenuse.
2. If the sides of a right triangle measure $$6$$ units and $$8$$ units respectively, find the length of the hypotenuse.
3. If the two equal sides of an isosceles right triangle measure $$7$$ units, find the length of the hypotenuse.
4. If the two equal sides of an isosceles right triangle measure $$10$$ units, find the length of the hypotenuse.
5. Calculates the diagonal of a square with sides $$3$$ centimeters.
6. Calculates the diagonal of a square with sides $$10$$ centimeters.
7. Calculates the diagonal of a square with side length $$\sqrt{6}$$ centimeters.
8. Calculates the diagonal of a square with sides $$\sqrt{10}$$ centimeters.
9. Calculates the length of the diagonal of a parallelogram with dimensions $$4$$ cm x $$8$$ cm.
10. Calculates the length of the diagonal of a rectangle with dimensions $$8$$ meters x $$10$$ meters.
11. Calculates the diagonal length of a rectangle with dimensions $$\sqrt{3}$$ meters x $$2$$ meters.
12. Calculates the diagonal length of a rectangle with dimensions $$\sqrt{6}$$ meters x $$\sqrt{10}$$ meters.
13. To ensure that the newly built door is square, the measured diagonal must match the distance calculated using the Pythagorean theorem. If the door measures $$4$$ feet by $$4$$ feet, how many inches should the diagonal measure? (Rounded to the nearest tenth of an inch.)
14. If the door frame measures $$3.5$$ feet by $$6.6$$ feet, what must be the diagonal dimension to ensure that the door frame is a perfect rectangle?

1. $$5$$ units

3. $$7\sqrt{2}$$ units

5. $$3\sqrt{2}$$ cm

7. $$2\sqrt{3}$$ cm

9. $$4\sqrt{5}$$ cm

11. $$\sqrt{7}$$ Rice

13. The diagonal measurement should be approximately "67.9" inches.

Exercise $$\PageIndex{7}$$

Determines whether the given triangle with sides a and b and hypotenuse c is a right triangle.

1. $$a = 3, b = 7,$$ and $$c = 10$$
2. $$a = 5, b = 12,$$ and $$c = 13$$
3. $$a = 8, b = 15,$$ and $$c = 17$$
4. $$a = 7, b = 24,$$ and $$c = 30$$
5. $$a = 3, b = 2,$$ and $$c = \sqrt{13}$$
6. $$a = \sqrt{7}, b = 4,$$ and $$c = \sqrt{11}$$
7. $$a = 4, b = \sqrt{3} ,$$ and $$c = \sqrt{19}$$
8. $$a = \sqrt{6} , b = \sqrt{15} and \(c = 21$$

1. Not a right triangle.

3. Right triangle.

5. Right triangle.

7. Right triangle.

Exercise $$\PageIndex{8}$$

1. What does the calculator display after calculating the square root of a negative number? Share your results on the discussion board and explain why you say this.
2. Study and discuss the history of the Pythagorean theorem.
3. Study and discuss the history of square roots.
4. Discuss the significance of the principal square root. Why cube root didn't work for the same problem? Give some examples and explain.

## footnote

74This number is multiplied by itself to get the original number.

75The symbol $$√$$ is used to represent the square root.

76Non-negative square root.

78The number is multiplied by itself three times to get the original number, denoted by $$\sqrt [ 3 ] { }$$.

79A positive integer $$n$$ in the notation $$\sqrt [ n ] { }$$ is used to denote the root.

80Given real numbers $$\sqrt [ n ] { A }$$ and $$\sqrt [ n ] { B }$$, $$\sqrt [ n ] { A \cdot B } = \sqrt [ n ] { A } \cdot \sqrt [ n ] { B }$$

81Given real numbers $$\sqrt [ n ] { A }$$ and $$\sqrt [ n ] { B }$$, $$\sqrt [ n ] { \frac { A } { B } } = \frac { \sqrt [ n ] { A } } { \sqrt [ n ] { B } }$$.

82A root in which the number contains no factors that can be written as perfect powers of exponents.

83Triangle with angle $$90°$$ .

84The longest side of a right triangle; is always the side opposite the right angle.

85The non hypotenuse side of a right triangle.

86The hypotenuse of any right triangle is equal to the square root of the sum of the squares of the lengths of the sides of the triangle.

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