##### Powers and Roots

An expression that represents repeated multiplication of the same factor is called a power. So, when we want to multiply a number by itself repeatedly we use powers.

$ \displaystyle {{a}^{n}}$ tells you to multiply a by itself, exactly n-times. The exponent corresponds to the number of times the base is used as a factor.

Worked example 1. Calculate:

a) 3²

b) 2³

c) $\displaystyle {{2}^{4}}$

**d) **$\displaystyle {{5}^{3}}$

Solution

a) 3² = 3 x 3 = 9

b) 2³ = 2 x 2 x 2 = 8

**c)** $\displaystyle {{2}^{4}}=2\cdot 2\cdot 2\cdot 2=16$

**d)** 5³ = 5 x 5 x 5 = 125

In words:

**3² **can be called: *“**3 **to the power 2**”*, or “*3 to the second power“*, or simply: *“3 squared“.*

**2³** can be called: *“2 to the third power”*, or *“2 to the power 3”*, or simply: *“2 cubed”*

**$\displaystyle {{2}^{4}}$** can be called: *“2 to the fourth power”*, or *“2 to the power 4”* or simply: *“2 to the fourth”*

Worked example 2

a) Write the product as a power: 3 x 3 x 3 x 3 ; 11 x 11 ; 7 x 7 x 7 x 7 x 7 x 7 x 7.

b) Evaluate each power : $\displaystyle {{3}^{6}}$ ; $\displaystyle {{9}^{3}}$; $\displaystyle {{10}^{2}}$

Solution

a) $\displaystyle 3\times 3\times 3\times 3={{3}^{4}}$

11 x 11 = 11²

$\displaystyle 7\times 7\times 7\times 7\times 7\times 7\times 7={{7}^{7}}$

b) $\displaystyle {{3}^{6}}=3\times 3\times 3\times 3\times 3\times 3$

9³ = 9 x 9 x 9

10² = 10 x 10

Square numbers and square roots

Square numbers or perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, the square of 2 × 2 = 4. The symbol for squared is ². So, 4 x 4 can also be written as 2².

### How do we find perfect squares?

As we sad a perfect square is an integer that is the square of an integer, so we create one by multiplying two equal integers by each others.

For example: 1 x 1 is a perfect square also 2 x 2 ………10 x 10 and so on.

We can also find a perfect square by finding its square root. If it’s a whole integer than it’s a perfect square.

For example: The square root of 36 is 6 and -6, since 6 and -+are whole numbers then 36 is a perfect square.

The square root of 37 is not a whole integer, so 37 it’s not a perfect square.

### The first 10 perfect squares

Multiply each of the first 10th numbers with each other. We can also square the negative numbers acting the same as we did above by multiplying the first 10th negative numbers.

### 1 x 1 = 1

2 x 2 = 4

3 x 3 = 9

4 x 4 = 16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81

10 x 10 = 100

(-1) x (-1) = 1

(-2 ) x (- 2 ) = 4

(-3) x (-3) = 9

(-4) x (-4) = 16

(-5) x (-5) = 25

(-6) x (-6) = 36

(-7) x (-7) = 49

(-8 ) x (-8) = 64

(-9) x (-9) = 81

(-10) x (-10) = 100

So, the first 10 perfect square numbers are: **1**, **4**, **9**, **16**, **25**, **36**, **49**, **64**, **81**, **100**.

## What do you notice?

We get the same result as when we multiplied the 10 first positive numbers and that’s because when we multiply too negative numbers we get a positive number.

The square root of a number is the number that was multiplied by itself to get the square number. The symbol for square is root is $\displaystyle \sqrt{{}}$.

You know that 25 = 5² . So, $\displaystyle \sqrt{25}=5$.

The square root of a number is the opposite of squaring a number.

The square root of “b” is a number “a” whose square is b.

a² = b

Be careful!

We found the square of 5 and -5 to be the same number 25.

Therefore, both 5 and -5 are square roots of 25 since the square root of a number is the opposite of squaring a number.

So, there can be two square roots of a number, a positive and a negative one.

$\displaystyle a=+\sqrt{b}$

$\displaystyle a=-\sqrt{b}$

Then why in textbooks we see it like $\displaystyle \sqrt{16}=4$ only?

Based on the “principal square root” or sometimes called “the positive square root” when we use the radical symbol we are going to use only the positive root.

Worked example 3: Find the square root of 36 and find $\displaystyle \sqrt{36}$

Solution: Based on what we explained above:

- The square root of 36 is 6 and -6
- The $\displaystyle \sqrt{36}=6$ only based on the “Principal square root”

### The square roots of all the perfect squares from 1 to 100

$\displaystyle \sqrt{1}$ = 1 since 1² = 1

$\displaystyle \sqrt{4}$ = 2 since 2² = 4

$\displaystyle \sqrt{9}$ = 3 since 3² = 9

$\displaystyle \sqrt{16}$ = 4 since 4² = 16

$\displaystyle \sqrt{25}$ = 5 since 5² = 25

$\displaystyle \sqrt{36}$ = 6 since 6² = 36

$\displaystyle \sqrt{49}$ = 7 since 7² = 49

$\displaystyle \sqrt{64}$ = 8 since 8² = 64

$\displaystyle \sqrt{81}$ = 9 since 9² = 81

$\displaystyle \sqrt{100}$ = 10 since 10² = 100

Worked example 4: Find which of the numbers in the brackets it’s a perfect square: ( 784, 202, 99, 23, 144)

Solution: If the square root of the number is a whole integer then the number it’s a perfect square.

$\displaystyle \sqrt{784}=28$, so, 28 it’s a whole number, so, 784 is a perfect square.

Without a calculator u can find if the number it’s perfect square by using Prime factorization

1) 784

784 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 7 ⋅ 7

784 = 24 ⋅ 72

$\displaystyle \sqrt{784}=\sqrt{{{2}^{4}}\cdot {{7}^{2}}}$

$\displaystyle \sqrt{784}=4\cdot 7=28$

So, we found that 784 it’s a perfect square.

2) 202

202 = 2 ⋅ 101

$\displaystyle \sqrt{202}=\sqrt{2}\cdot \sqrt{101}$

So, 202 is not a perfect square.

3) 99

9 = 3 ⋅ 3 ⋅ 11

$\displaystyle \sqrt{99}=\sqrt{3\cdot 3\cdot 11}$

$\displaystyle \sqrt{99}=\sqrt{{{3}^{^{2}}}}\cdot \sqrt{11}$

$\displaystyle \sqrt{99}=3\cdot \sqrt{11}$

So, 99 is not a perfect square.

4) 23

23 = 1 ⋅ 23

$\displaystyle \sqrt{23}=\sqrt{1\cdot 23}$

Therefore 23 is not a perfect square.

5) 144

144 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3

144 = 24 ⋅ 32

$ \displaystyle \sqrt{144}=\sqrt{{{2}^{4}}\cdot {{3}^{2}}}$

$\displaystyle \sqrt{144}=4\cdot 3$

$\displaystyle \sqrt{144}=12$

So, 144 is a perfect square.

### What about finding the square root of large numbers and numbers that aren’t perfect squares?

The easiest way it’s using a calculator, but what if we don’t have one?

We can use The Heron’s Method:

Step 1: Find the ‘’a” closest square root to our “b”non-perfect square.

Step 2: Use formula $\displaystyle \frac{1}{2}\left( a\text{ }+\text{ }\frac{b}{a} \right)$

Step 3: Repeat the formula.

**Worked example 5**: Find a square root of 115.

**Solution:**

Step 1: Find the closest square root to our non-perfect square 115.

The closest perfect square is a = 11 because

$\displaystyle {{11}^{^{2}}}=121$

a = 11

b = 115

Step 2: Use the formula:

$\displaystyle \frac{1}{2}\left( a+\frac{b}{a} \right)$

$\displaystyle \frac{1}{2}(11+\frac{115}{11})=$

$ \displaystyle \frac{1}{2}(\frac{121+115}{11})=$

$\displaystyle \frac{1}{2}(\frac{236}{11})=$

$\displaystyle \frac{236}{22}=10.\overline{72}$

Step 3: Repeat the formula

Now the $\displaystyle a=10.\overline{72}$

b = 115

$\displaystyle \frac{1}{2}\left( a+\frac{b}{a} \right)=$

$\displaystyle \frac{1}{2}(10.\overline{72}+\frac{115}{10.\overline{72}})=$

$\displaystyle \frac{1}{2}(\frac{10.\overline{72}+115}{10.\overline{72}})=$

$\displaystyle =10.7238\ldots $

A square root of 115 is 10.7238

### Square numbers and cube root

A number is cubed when it is multiplied by itself and then multiplied by itself again.For example, the cube of 2 is 2 x 2 x 2 = 8. The symbol for cubed is ³. So, 2 x 2 x 2 can also written as 2³

The cube root of a number is a number that was multiplied three times by itself to get the cube number. The symbol for cube root is $\displaystyle \sqrt[3]{{}}$ . You know that 8 = 2³ , so $\displaystyle \sqrt[3]{8}=2$.