## Section1.3Absolute Value and Square Root

¶In this section, we will learn the basics of absolute value and square root. These are actions you can *do* to a given number, often changing the number into something else.

### Subsection1.3.1Introduction to Absolute Value

###### Definition1.3.2

The absolute value of a number is the distance between that number and \(0\) on a number line. For the absolute value of \(x\text{,}\) we write \(\abs{x}\text{.}\)

Let's look at \(\abs{2}\) and \(\abs{-2}\text{,}\) the absolute value of \(2\) and \(-2\text{.}\)

Since the distance between \(2\) and \(0\) on the number line is \(2\) units, the absolute value of \(2\) is \(2\text{.}\) We write \(\abs{2}=2\text{.}\)

Since the distance between \(-2\) and \(0\) on the number line is also \(2\) units, the absolute value of \(-2\) is also \(2\text{.}\) We write \(\abs{-2}=2\text{.}\)

###### Fact1.3.4Absolute Value

Taking the absolute value of a number results in whatever the “positive version” of that number is. This is because the real meaning of absolute value is its *distance* from zero.

###### Checkpoint1.3.5Calculating Absolute Value

Try calculating some absolute values.

###### Warning1.3.6Absolute Value Does Not Exactly “Make Everything Positive”

Students may see an expression like \(\abs{2-5}\) and incorrectly think it is OK to “make everything positive” and write \(2+5\text{.}\) This is incorrect since \(\abs{2-5}\) works out to be \(3\text{,}\) not \(7\text{,}\) as we are actually taking the absolute value of \(-3\) (the equivalent number inside the absolute value).

### Subsection1.3.2Square Root Facts

If you have learned your basic multiplication table, you know:

\(\times\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |

\(1\) | \(1\) | \(\lowlight{2}\) | \(\lowlight{3}\) | \(\lowlight{4}\) | \(\lowlight{5}\) | \(\lowlight{6}\) | \(\lowlight{7}\) | \(\lowlight{8}\) | \(\lowlight{9}\) |

\(2\) | \(\lowlight{2}\) | \(4\) | \(\lowlight{6}\) | \(\lowlight{8}\) | \(\lowlight{10}\) | \(\lowlight{12}\) | \(\lowlight{14}\) | \(\lowlight{16}\) | \(\lowlight{18}\) |

\(3\) | \(\lowlight{3}\) | \(\lowlight{6}\) | \(9\) | \(\lowlight{12}\) | \(\lowlight{15}\) | \(\lowlight{18}\) | \(\lowlight{21}\) | \(\lowlight{24}\) | \(\lowlight{27}\) |

\(4\) | \(\lowlight{4}\) | \(\lowlight{8}\) | \(\lowlight{12}\) | \(16\) | \(\lowlight{20}\) | \(\lowlight{24}\) | \(\lowlight{28}\) | \(\lowlight{32}\) | \(\lowlight{36}\) |

\(5\) | \(\lowlight{5}\) | \(\lowlight{10}\) | \(\lowlight{15}\) | \(\lowlight{20}\) | \(25\) | \(\lowlight{30}\) | \(\lowlight{35}\) | \(\lowlight{40}\) | \(\lowlight{45}\) |

\(6\) | \(\lowlight{6}\) | \(\lowlight{12}\) | \(\lowlight{18}\) | \(\lowlight{24}\) | \(\lowlight{30}\) | \(36\) | \(\lowlight{42}\) | \(\lowlight{48}\) | \(\lowlight{54}\) |

\(7\) | \(\lowlight{7}\) | \(\lowlight{14}\) | \(\lowlight{21}\) | \(\lowlight{28}\) | \(\lowlight{35}\) | \(\lowlight{42}\) | \(49\) | \(\lowlight{56}\) | \(\lowlight{63}\) |

\(8\) | \(\lowlight{8}\) | \(\lowlight{16}\) | \(\lowlight{24}\) | \(\lowlight{32}\) | \(\lowlight{40}\) | \(\lowlight{48}\) | \(\lowlight{56}\) | \(64\) | \(\lowlight{72}\) |

\(9\) | \(\lowlight{9}\) | \(\lowlight{18}\) | \(\lowlight{27}\) | \(\lowlight{36}\) | \(\lowlight{45}\) | \(\lowlight{54}\) | \(\lowlight{63}\) | \(\lowlight{72}\) | \(81\) |

The numbers along the diagonal are special; they are known as perfect squares. And for working with square roots, it will be helpful if you can memorize these first few perfect square numbers.

“Taking a square root” is the opposite action of squaring a number. For example, when you square \(3\text{,}\) the result is \(9\text{.}\) So when you take the square root of \(9\text{,}\) the result is \(3\text{.}\) Just knowing that \(9\) comes about as \(3^2\) lets us realize that \(3\) is the square root of \(9\text{.}\) This is why memorizing the perfect squares from the multiplication table can be so helpful.

The notation we use for taking a square root is the radical, \(\sqrt{\phantom{x}}\text{.}\) For example, “the square root of \(9\)” is denoted \(\sqrt{9}\text{.}\) And now we know enough to be able to write \(\sqrt{9}=3\text{.}\)

Tossing in a few extra special square roots, it's advisable to memorize the following:

\(\sqrt{0}=0\) | \(\sqrt{1}=1\) | \(\sqrt{4}=2\) | \(\sqrt{9}=3\) |

\(\sqrt{16}=4\) | \(\sqrt{25}=5\) | \(\sqrt{36}=6\) | \(\sqrt{49}=7\) |

\(\sqrt{64}=8\) | \(\sqrt{81}=9\) | \(\sqrt{100}=10\) | \(\sqrt{121}=11\) |

\(\sqrt{144}=12\) | \(\sqrt{169}=13\) | \(\sqrt{196}=14\) | \(\sqrt{225}=15\) |

### Subsection1.3.3Calculating Square Roots with a Calculator

Most square roots are actually numbers with decimal places that go on forever. Take \(\sqrt{5}\) as an example:

Since \(5\) is between \(4\) and \(9\text{,}\) then \(\sqrt{5}\) must be somewhere between \(2\) and \(3\text{.}\) There are no whole numbers between \(2\) and \(3\text{,}\) so \(\sqrt{5}\) must be some number with decimal places. If the decimal places eventually stopped, then squaring it would give you another number with decimal places that stop further out. But squaring it gives you \(5\) with no decimal places. So the only possibility is that \(\sqrt{5}\) is a decimal between \(2\) and \(3\) that goes on forever. With a calculator, we can see

Actually the decimal will not terminate, and that is why we used the \(\approx\) symbol instead of an equals sign. To get \(2.236\) we rounded down slightly from the true value of \(\sqrt{5}\text{.}\) With a calculator, we can check that \(2.236^2=4.999696\text{,}\) a little shy of \(5\text{.}\)

### Subsection1.3.4Square Roots of Fractions

We can calculate the square root of some fractions by hand, such as \(\sqrt{\frac{1}{4}}\text{.}\) The idea is the same: can you think of a number that you would square to get \(\frac{1}{4}\text{?}\) Being familiar with fraction multiplication, we know that

and so \(\sqrt{\frac{1}{4}}=\frac{1}{2}\text{.}\)

###### Checkpoint1.3.8Square Roots of Fractions

Try calculating some absolute values.

### Subsection1.3.5Square Root of Negative Numbers

Can we find the square root of a negative number, such as \(\sqrt{-25}\text{?}\) That would mean that there is some number out there that multiplies by itself to make \(-25\text{.}\) Would \(\sqrt{-25}\) be positive or negative? Either way, once you square it (multiply it by itself) the result would be positive. So it couldn't possibly square to \(-25\text{.}\) So there is no square root of \(-25\) or of any negative number for that matter.

If you are confronted with an expression like \(\sqrt{-25}\text{,}\) or any other square root of a negative number, you can state that “there is no real square root” or that the result “does not exist” (as a real number).

### SubsectionExercises

These skills practice familiarity with absolute value.

###### 1

Find the absolute value of this number.

\(\displaystyle{ |{-7}|= }\)

###### 2

Find the absolute value of this number.

\(\displaystyle{ |{-5}|= }\)

###### 3

Find the absolute value of the following numbers.

\(\displaystyle{ |{5}|= }\)

\(\displaystyle{ |{-5}|= }\)

\(\displaystyle{ -|{5}|= }\)

\(\displaystyle{ -|{-5}|= }\)

###### 4

Find the absolute value of the following numbers.

\(\displaystyle{ |{6}|= }\)

\(\displaystyle{ |{-6}|= }\)

\(\displaystyle{ -|{6}|= }\)

\(\displaystyle{ -|{-6}|= }\)

###### 5

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ \left| 7 \right| = }\)

\(\displaystyle{ \left| -1 \right| = }\)

\(\displaystyle{ \left| 0 \right| = }\)

\(\displaystyle{ \left| {13+\left(-4\right)} \right| = }\)

\(\displaystyle{ \left| {-5-\left(-3\right)} \right| = }\)

###### 6

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ \left| 8 \right| = }\)

\(\displaystyle{ \left| -4 \right| = }\)

\(\displaystyle{ \left| 0 \right| = }\)

\(\displaystyle{ \left| {17+\left(-7\right)} \right| = }\)

\(\displaystyle{ \left| {-5-\left(-1\right)} \right| = }\)

###### 7

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ - \lvert 2-6 \rvert = }\)

\(\displaystyle{ \lvert -2-6 \rvert = }\)

\(\displaystyle{ -2 \lvert 6-2 \rvert = }\)

###### 8

Evaluate the following expressions which involve the absolute value:

\(\displaystyle{ - \lvert 5-8 \rvert = }\)

\(\displaystyle{ \lvert -5-8 \rvert = }\)

\(\displaystyle{ -2 \lvert 8-5 \rvert = }\)

These skills practice familiarity with square roots.

###### 9

Which of the following are square numbers? There may be more than one correct answer.

\(73\)

\(1\)

\(129\)

\(144\)

\(66\)

\(64\)

###### 10

Which of the following are square numbers? There may be more than one correct answer.

\(34\)

\(36\)

\(16\)

\(137\)

\(4\)

\(127\)

###### 11

Find the square root of the following numbers:

\(\displaystyle{ \sqrt{16} }\) =

\(\displaystyle{ \sqrt{121} }\) =

\(\displaystyle{ \sqrt{81} }\) =

###### 12

Find the square root of the following numbers:

\(\displaystyle{ \sqrt{25} }\) =

\(\displaystyle{ \sqrt{49} }\) =

\(\displaystyle{ \sqrt{9} }\) =

###### 13

Find the square root of the following numbers.

\(\displaystyle{ \sqrt{{{\frac{36}{25}}}} }\) =

\(\displaystyle{ \sqrt{{-{\frac{121}{16}}}} }\) =

###### 14

Find the square root of the following numbers.

\(\displaystyle{ \sqrt{{{\frac{64}{121}}}} }\) =

\(\displaystyle{ \sqrt{{-{\frac{1}{9}}}} }\) =

###### 15

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{81} }\) =

\(\displaystyle{ \sqrt{0.81} }\) =

\(\displaystyle{ \sqrt{8100} }\) =

###### 16

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{121} }\) =

\(\displaystyle{ \sqrt{1.21} }\) =

\(\displaystyle{ \sqrt{12100} }\) =

###### 17

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{144} }\) =

\(\displaystyle{ \sqrt{14400} }\) =

\(\displaystyle{ \sqrt{1440000} }\) =

###### 18

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{1} }\) =

\(\displaystyle{ \sqrt{100} }\) =

\(\displaystyle{ \sqrt{10000} }\) =

###### 19

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{4} }\) =

\(\displaystyle{ \sqrt{0.04} }\) =

\(\displaystyle{ \sqrt{0.0004} }\) =

###### 20

Find the square root of the following numbers without using a calculator.

\(\displaystyle{ \sqrt{16} }\) =

\(\displaystyle{ \sqrt{0.16} }\) =

\(\displaystyle{ \sqrt{0.0016} }\) =

###### 21

Use a calculator to approximate a decimal value for \(\sqrt{38}\text{.}\)

###### 22

Use a calculator to approximate a decimal value for \(\sqrt{48}\text{.}\)

###### 23

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{\frac{49}{81}}\) is .

###### 24

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{\frac{64}{81}}\) is .

###### 25

Simplify the radical expression or state that it is `not a real number`

.

\(-\sqrt{121}\) is .

###### 26

Simplify the radical expression or state that it is `not a real number`

.

\(-\sqrt{144}\) is .

###### 27

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{-4}\) is .

###### 28

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{-9}\) is .

###### 29

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{-{{\frac{9}{64}}}}\) is .

###### 30

Simplify the radical expression or state that it is `not a real number`

.

\(\sqrt{-{{\frac{25}{81}}}}\) is .

###### 31

Simplify the radical expression or state that it is `not a real number`

.

\(-\sqrt{{{\frac{36}{121}}}}\) is .

###### 32

Simplify the radical expression or state that it is `not a real number`

.

\(-\sqrt{{{\frac{49}{64}}}}\) is .

###### 33

Simplify the radical expression or state that it is `not a real number`

.

\(\displaystyle{\sqrt{169}-\sqrt{25}=}\)

\(\displaystyle{\sqrt{169-25}=}\)

###### 34

Simplify the radical expression or state that it is `not a real number`

.

\(\displaystyle{\sqrt{25}-\sqrt{9}=}\)

\(\displaystyle{\sqrt{25-9}=}\)

###### 35

Simplify the expression so that there is no longer a radical in the denominator.

\(\displaystyle{\frac{9}{\sqrt{{16}}}}\) = .

###### 36

Simplify the expression so that there is no longer a radical in the denominator.

\(\displaystyle{\frac{1}{\sqrt{{9}}}}\) = .