## Wednesday, May 9, 2012

Here's part of chapter 2 from my second book, Algebra Simplified Intermediate & Advanced. If interested in the book, go to http://www.lulu.com/spotlight/KKauffman1969

# Chapter 2

## Square Roots, Cube Roots and Higher Roots

Recall when raising a number to a power n, where n is an integer greater than 1, we multiply the number by itself n times. For example, 43 = 4 ∙ 4 ∙ 4. Now suppose we want to know what number multiplied by itself 2 times equals 169. Problems of this kind can be represented using radicals.

A radical symbol √ is used to show the square root, or principal square root of a number or expression that appears under the radical symbol. Recall that the square root is defined as a number or expression multiplied by itself twice to equal the number or expression under the radical symbol, known as the radicand.

For example, if we want to know what number multiplied by itself 2 times equals 169, we can set this up with the radical symbol as follows:

169, read as “square root of 169”. The answer to this is 13.

- √169, read as “negative square root of 169”. The answer to this is -13.

0.09 = 0.3 and -0.3 since (0.3)2 and (-0.3)2 equals 0.09. The principal square root is 0.3. Another way to simplify this is to change √0.09 to √(9/100) and simplify to 3/10.

(25/49) = 5/7 and -5/7 since (5/7)2 and (-5/7)2. The principal square root is 5/7.

Note that a square root also has a negative value since a negative times a negative equals a positive, but we will deal with only the principal square root unless otherwise noted.

Note that you can also simplify the square root of a fraction by taking the square root of the numerator and then the square root of the denominator instead of the square root of the fraction as a whole. In the previous example, you can take the square root of 25 first, then the square root of 49.

Note that the square root of many positive integers are not whole numbers or rational numbers. For example, √19 can be found on a calculator or by leaving the answer as √19.

An easy way to solve many square root problems is to know the perfect squares from 1 to 25. They are as follows:

12 = 1 102 = 100 192 = 361

22 = 4 112 = 121 202 = 400

32 = 9 122 = 144 212 = 441

42 = 16 132 = 169 222 = 484

52 = 25 142 = 196 232 = 529

62 = 36 152 = 225 242 = 576

72 = 49 162 = 256 252 = 625.

82 = 64 172 = 289

92 = 81 182 = 324

Sometimes we have to find the square root of a number that is not a perfect square. In these cases, we break down the radicand into factors, one of which is a perfect square.

Examples: Find each square root.

1. √68

First, find factors of 68.

Since 68 is even, we can divide it by 2. Therefore, 68 = 2 ∙ 34. Notice 34 is also divisible by 2, therefore 34 = 2 ∙ 17.

So 68 is factored into 2 ∙ 2 ∙ 17. Notice that 17 is prime and cannot be factored further and 2 ∙ 2 = 4, which is a perfect square. Therefore √68 = √4 ∙ √17 = 2√17.

2. √108

First, find factors of 108.

Since 108 is even, we can divide it by 2. Therefore 108 = 2 ∙ 54. Notice 54 is also divisible by 2, therefore 54 = 2 ∙ 27. Next, we know that 27 = 3 ∙ 3 ∙ 3.

The factors of 108 are 2 ∙ 2 ∙ 3 ∙ 3 ∙ 3. Notice 2 ∙ 2 = 4, which is a perfect square and 3 ∙ 3 = 9, which is also a perfect square. Therefore, 108 = 4 ∙ 9 ∙ 3 and √108 = √4 ∙ √9 ∙ √3 = 2 ∙ 3 ∙ √3 = 6√3.

Note that 108 = 36 ∙ 3 and 36 is a perfect square. But if you can't see right away that 3 is a factor of 108, you can break down by dividing 108 by 2 first and then simplify further at that point. It's easy to determine that 108 is divisible by 3. If the sum of the digits of a number are divisible by 3, the number is divisible by 3.

Recall, for a variable x, where x ≠ 0, √x2 = x and -x. If we are dealing with just the principal square root, then the answer is |x|. Recall that |x| is the absolute value of x.

Examples: Find the square root of the following.

a. √(25x2)

Simplify the numerical part and the variable part separately.

25 = 5 since 52 = 25 and √x2 = x since (x)2 = x2.

Therefore, √(25x2) = 5x.

b. √(64x4y6)

64 = 8 since 82 = 64, √x4 = x2 since (x2)2 = x4 and √y6 = y3 since (y3)2 = y6.

Therefore, multiply the terms in bold to get √(64x4y6) = 8x2y3.

c. √(x + 4)2 = (x + 4) because (x + 4)(x + 4) = (x + 4)2

Note that the square root of any number or expression squared is just that number. For example, notice in the previous examples that √(x + 4)2 = (x + 4) and √25 = √(5)2 = 5.

d. √(x2 + 12x + 36) = √(x + 6)(x + 6)

Notice that we factored (x2 + 12x + 36) to get (x + 6)(x + 6). That is a perfect square which enables us to simplify the radical expression.

(x2 + 12x + 36) = √(x + 6)(x + 6) = √(x + 6)2 = (x + 6)

Note that you can check by squaring your answer. After squaring, you should get the expression under the √.

In the previous examples, we simplified radicals that were perfect squares. Many times we can simplify radicals that are not perfect squares. The idea is to break down the number or variable into factors, one of which is a perfect square. By the  multiplication property of radicals n√(ab) = na nb, where n√a and nb are real numbers.

Note that in a square root n is 2 but is not written. When solving any root higher than 2, it is noted in the upper left of the radical. For example, the cube root (n = 3) of a is noted as 3√a.

The cube root of a number n is the number when multiplied by itself 3 times equals n. For example, the cube root of -125 is -5 because (-5) ∙ (-5) ∙ (-5 ) = -125, or (-5)3 = -125. By definition, the cube root of a is defined as 3a = b, if b3 = a.

Examples:

3√27 = 3 since 33 = 27

3√(-27) = -3 since (-3)3 = -27

3√125 = 5 since 53 = 125

3√(-125) = -5 since (-5)3 = -125

Note that 27 and 125 have 1 real number cube root and 2 real number square roots. Any negative number will have a cube root that is negative and any positive number will have a cube root that is positive.

Examples: Simplify the following.

a. 3√-8

To solve this, think of what number multiplied by itself 3 times gives you -8. Notice that (-2) ∙ (-2) ∙ (-2) = -8. Therefore, 3√-8 = -2.

b. 3√343 = 7 since 7 ∙ 7 ∙ 7 = 343.

c. 3√729 = -9

Note that if the sum of the digits of a number is divisible by 9, the number is divisible by 9.

The sum of the digits of 729 is 18, which is divisible by 9. Therefore, 729 is divisible by 9 and -729 is divisible by 9.

729/9 = 81 and we know the square root of 81 is 9.

We reviewed how to take the cube root of positive and negative numbers. Now we will review how to take the cube root involving variables. Recall that by definition, if x is a variable representing a real number, 3x3 = x.

Examples: Simplify the following.

a. 3√(27x3) = 3√(27) ∙ 3√(x3) = 3 ∙ x = 3x

Check by taking the cube of 3x.

(3x)3 = 3x ∙ 3x ∙ 3x = 27x3.

b. 3√(5x + 2y)3 = (5x + 2y) because of the rule that 3x3 = x.

c. 3√(-216x6y12)

3√(-216) = -6, 3√(x6) = x2 since (x2)3 = x6 and 3√(y12) = y4 since (y4)3 = y12.

Multiply the terms in bold to get -6x2y4.

The previous examples were perfect cubes, but oftentimes we can find the cube root of expressions that are not perfect cubes. The idea is to break down the number or variable into factors, one of which is a perfect cube. For example, if you solve 3√81, break it down into 3√27 ∙ 3√3 because 3√27 = 3. If you solve 3√(y5), break it down into 3√(y3) ∙ 3√(y2) because 3√(y3) = y. Remember from the multiplication property of radicals that 3ab = 3a3b.

Example: Simplify the following.

3√(16x4)

3√16 = 3√8 ∙ 3√2 = 23√2 (3√8 = 2 because 23 = 8)

3√(x4) = 3√(x3) ∙ 3x = x3x (3√(x3) = x by the definition of cube root (x)3 = x3)

Multiply like terms to get 3√(16x4) = 2x3√(2x)

Note that you can check by cubing the answer, [2x3√(2x)]3 = 8x3 ∙ 2x = 16x4.

Example: Simply the following.

3√(-192x4y5)

3√(-192) = 3√(-64) ∙ 3√(3) = -43√(3) (note that -64 ∙ 3 = -192 and (-4)3 = -64)

3√(x4) = 3√(x3) ∙ 3√(x) = x3√(x) (note that x3x = x4 and (x)3 = x3)

3√(y5) = 3√(y3) ∙ 3√(y2) = y 3√(y2) (note that y3y2 = y5 and (y)3 = y3)

Now multiply the like terms to get -4xy3√(3xy2).

In some cases cube roots will be the form of a fraction of two numbers, a number and a expression or two expressions. In these cases, try to simplify the ratio first. By the division property of radicals 3√(x/y) = 3x / 3y.

Example: Simplify the following.

3√(8x3/64y6)

By the division property of radicals 3√(8x3/64y6) = 3√(8x3) / 3√(64y6)

3√(8x3) = 2x because (2x)3 = 8x3

3√(64y6) = 4y2 because (4y2)3 = 64y6

Therefore 3√(8x3/64y6) = (2x)/(4y2) = x/(2y2).

Example: Simplify the following.

3√(216y5/8y8)

By the division property of radicals 3√(216y5/8y8) = 3√(27/y3) because 216/8 = 27 and (y5)/(y8) = 1/(y3).

3√(27) = 3 and 3√(1/y3) = 1/y.

Therefore 3√(216y5/8y8) = 3/y.

Note that it's easier to solve cube roots if you know the first several perfect cubes. The first ten perfect cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000.

We reviewed how to take the square root and cube root of numbers and expressions. There are also fourth roots, fifth roots, sixth roots, seventh roots and so on. The procedure for solving these roots is the same as for square roots and cube roots. For example, a fourth root of a natural number x is a number multiplied by itself four times to equal x, and so on for higher roots. If n is an odd natural number greater than 1 (n > 1), then n√(x) is an odd root. When n is an even natural number greater than 1 (n > 1) and x > 0, then n√(x) is an even root.

Examples: Simplify the following radical expressions.

a. 4√10000

4√10000 = 10 since 104 = 10 ∙ 10 ∙ 10 ∙ 10 = 10000

b. 5√(32x15)

5√32 = 2 because 25 = 32

5√(x15) = x3 because (x3)5 = x15

Therefore 5√(32x15) = 2x3

c. 6√(729x6y18)

6√729 = 3 because 36 = 729

6√(x6) = x because (x)6 = x6

6√(y18) = y3 because (y3)6 = y18

Therefore 6√(729x6y18) = 3xy3

d. 7√(x + 10)14

7√(x + 10)14 = (x + 10)2 because [(x + 10)2]7 = (x + 10)14

### Review Problems: Set 1

Simplify each of the following square roots.

1. √121 2. √(4/25) 3. -√0.49

4. √60 5. √125 6. √252

Simplify each of the following radical expressions.

7. √(9y2) 8. √(64x4) 9. √(x2 + 14x + 49)

10. √(y2 – 8y + 16) 11. √(81x8y6) 12. √(144x4y2z6)

13. √(80x5) 14. √(75xy2z3) 15. √(98z3)

Simplify each of the following cube roots.

16. 3√-64 17. 3√(27/125) 18. -3√-216

19. 3√40 20. 3√320 21. 3√-54

Simplify each of the following radical expressions.

22. 3√(8x4y5z6) 23. 3√(-108z4) 24. 4√256

25. 5√(-243a5) 26. 6√(32x12) 27. 7√(107)

### Rational Exponents

Just as we can raise a number or expression to exponents that are integers, we can raise them to fractional exponents as well. If n is a positive integer greater then 1, then x1/n is the nth root of x and 1/n is a rational exponent. In other words x1/n = nx.

To make this more clear, consider √3 and 31/2. By squaring both of these, we can determine that they are equal.

(√3)2 = (√3)(√3) = √9 = 3

(31/2)2 = 3(1/2 ∙ 2) = 31 = 3

Notice that squaring both gives a result of 3. Therefore √3 = 31/2.

We can rewrite exponential expressions in radical form and radical expressions in exponential form. The radicand of a radical expression becomes the base of an exponential expression and one over the index of the radical expression becomes the exponent. We'll illustrate this in the following

36 = 6 1/3. Notice the index in red becoming the denominator of the exponent and the radicand in blue becomes the base.

Examples: Evaluate the following.

1. 161/2 = √16 = 4. Recall that the index of a square root is 2, although the 2 is not written.

2. (-27)1/3 = 3√(-27) = -3 because (-3)3 = -27.

3. (16/81)1/4 = 4√(16/81) = 2/3 because (2/3)4 = 16/81.

4. -321/5 = -2 because (-2)5 = -32.

Examples: Simplify the following rational expressions.

1. (25x4)1/2

We will break this down into the number part and the variable part.

251/2 = 5 and (x4)1/2 = √(x4) = x2 since (x2)2 = x4.

Therefore (25x4)1/2 = 5x2.

Note that (x4)1/2 can also be simplified using the rules for raising an exponent to an exponent. Therefore, the exponents 4 and ½ are multiplied together to get 2, which is the new exponent and (x4)1/2 = x2.

2. (-125x3)1/3

(-125)1/3 = 3√(-125) = -5 since (-5)3 = -125 (x3)1/3 = 3√(x3) = x

Therefore, (-125x3)1/3 = -5x.

3. [(2a + 7)4]1/4 = 4√(2a + 7)4 = |2a + 7|

Notice the answer is |2a + 7| and not 2a + 7. |2a + 7|4 = (2a + 7)4 and a can be any number, therefore 2a + 7 could be a negative number. But we can't have a real number solution if 2a + 7 is negative because no real number raised to the 4th power is negative. Therefore we must have absolute value to ensure a positive real number solution.

4. [(4b – 3)5]1/5 = 5√(4b – 3)5 = 4b – 3.

In this case, since n is odd, there is no need for the absolute value symbols in the solution. The value of 4b- 3 could still be negative but a negative quantity raised to an odd power will be negative.

In the previous examples all of the exponents were in the form 1/n. Now we will learn how to simplify exponential expressions where the exponent is in the form p/n where p and n are positive integers and n ≠ 1. To illustrate how to simplify , consider the following example.

272/3. We can rewrite this in the form x1/n raised to the 2nd power, (x1/n)2 = x2/n.

Therefore, 272/3 = (271/3)2

= (3√27)2 = 32 = 9

We can rewrite the exponential expression in radical form. The numerator of the rational exponent becomes the power of the radical expression. The denominator of the rational exponent becomes the index of the radical expression and the base of the exponential expression becomes the radicand. In the previous example,

272/3 = (327)2

We would solve this as follows:

Take the cube root of 27: 3√27 = 3

Square 3: 32 = 9

Or we can think of 272/3 as

272/3 = 3√(27)2

We would solve this as follows:

Square 27 first : 272 = 729

Take the cube root of 729 : 3√729 = 9 since 93 = 729.

Note that we know that 729 is divisible by 9 because the sum of the digits in 729 is divisible by 9. The sum of the digits of 729 is 18, which is divisible by 9.