MATH 2 part 1

C. Simplify the following rational expressions and express your answer in lowest terms.1. _4a2 – 20a 4a(a – 5)__ 4a__ a2 – 4a – 5 (a – 5)(a + 1) a + 12. _ m2 – 4___ (m – 2)(m + 2) m – 2 m2 + 4m + 4 (m + 2)(m + 2) m + 23. 12 – 7x + x2 (– 4 + x)( –3 + x) (– 3 + x) – 1 Take note (–3 + x) = (x + 3) (x – 3)(4 + x) –(–4 + x)(x – 3) – (x – 3)4. _ n2_– p2 _ (n + p)(n – p) n + p Take note – (n + p) n+p –1p–n – (n – p)5. _r 2 – r – 20 (r – 5)(r + 4) r + 4 r2 + r – 30 (r – 5)(r + 6) r + 66. _g3 + c3 _ (g + c)(g2 – cg + c2) (g2 – cg + c2) g2 – c2 (g + c) (g – c) g–c7. _ a2 + ac – ab – bc a(a + c) – b(a + c) (a – b)(a + c) a + c a2 – b2 (a – b)(a + b) (a – b)(a + b) a + b8. _m2 – mp + mn – np _m (m – p) + n(m – p) (m + n)(m – p) m – p m2 + 2mn + n2 (m + n)(m + n) (m + n)(m + n) m + n9. _xy – yw + xz – zw y(x – w) + z(x – w) (y + z)(x – w) x – w xy + yw + zx + zw y(x + w) + z(x + w) (y + z)(x + w) x + w10. ac + ad – bc – bd a(c + d) – b(c + d) (a – b)(c + d) c+d ab + ac – b2 – bc a(b + c) – b(b + c) (a – b)(b + c) b+cWhat have you learned1. 4x2 – 16 = 0 4x2 = 16 x = 2; –22. a2 + 7a + 12 = 0 (a + 3)(a + 4)=0 a = –3; –4 Domain = { a Є R a ≠ –3; –4}3. m2 – m – 12 = 0 (m – 4)(m + 3) = 0 m = -3; 4 25

4. x2 – 2x – 24 = 0 (x – 6)(x + 4) = 0 x = 6; – 4 Domain = { x Є R x ≠ 6; –4}5. 4x2 – 64 = 0 4x2 = 64 x = 4; – 46. 2(-3)2 – 5(-3) + 3 _2(9) +15 + 3_ 36 _9_ 3(-3)2 – 5(-3) + 2 3(9) + 15 + 2 44 117. 7(-2)2(-1)3 _ 7(4)(-1) –28 13 –3(-2)2 + 4(-1) –12 – 4 –16 48. _ - m7n4p8 _ __n2p2_ -18m12n2p6 18m59. _y4 – 13y2 + 36_ _(y2 – 9)(y2 – 4) _(y – 3)(y + 3) (y + 2) (y – 2) (y – 3)(y – 2)y2 + 5y + 6 (y + 2)(y + 3) (y + 3) (y + 2)10. _x4 – 16 _ _(x2 + 4)(x2 – 4) _ x2 + 4____ x4– 8x2 + 16 (x2 – 4)(x2 – 4) (x – 2)(x + 2) 26

Module 1Searching for Patterns in Sequences, Arithmetic, Geometric and Others What this module is all about This module will teach you how to deal with a lot of number patterns. Thesenumber patterns are called sequences. Go over the lessons and have fun in workingwith the exercises. What you are expected to learn It is expected that you will be able to demonstrate knowledge and skill relatedto sequences and apply these in solving problems. Specifically, you should be ableto: a) list the next few terms of a sequence given several consecutive terms. b) derive by pattern searching, a mathematical expression (rule) for generating the sequences. c) generate the next few terms of sequences defined recursively. d) describe an arithmetic sequence by any of the following: • giving the first few terms • giving the formula for the nth term • drawing the graphHow much do you knowA. Write the first five terms of the sequence.1. an = n + 3 3. an = 4n – 3 n2. an = 2n - 1B. Find the indicated term for the sequence.4. an = -9n + 2; a8 6. an = 3n + 7 ; a14 2n − 5

5. an = (n + 1)(2n + 3); a5C. Find the general term, an, for the given terms of the sequence.7. 4, 8, 12, 16,… 9. 1 , 1 , 1 ,...8. -10, -20, -30, -40,… 2 6 12D. Find the first four terms of the sequence defined recursively.10. a1= -1, an = 3an – 1 11. a1 = 5, an = an−1 n −1 E. Find the common difference and the next three terms of the givenarithmetic sequence.12. 2, 8, 14, 20,… 14. 1, 2, 3, 4,…13. 7, 8.5, 10, 11.5,… 15. 24, 21, 18, … What you will do Lesson 1 Sequences and its Kinds It is a common experience to be confronted with a set of numbers arranged insome order. The order and arrangement may be given to you or you have to discovera rule for it from some data. For example, the milkman comes every other day. Hecame on July 17; will he come on Aug 12? Consider that you are given the set ofdates 17, 19, 21,…arranged from left to right in the order of increasing time. Continuing the set we have 17, 19, 21, …, 29, 31, 2, 4, ….,28, 30…so that the answer to our question is yes. Any such ordered arrangement of a set of numbers is called a SEQUENCE. 2

Look at this second example. Lorna, a 2nd year student in a certain publicschool, is able to save the money her ninongs and ninangs gave her last Christmas.She then deposits her savings of P1,000 in an account that earns 10% simpleinterest. The total amount of interest she earned in each of the 1st 4 years of hersaving is shown below:Year 1234Total amount 10 20 30 40 The list of numbers 10, 20, 30, 40 is called a sequence. The list 10,20,30,40is ordered because the position in this list indicates the year in which that totalamount of interest is earned. Now, each of the numbers of a sequence is called a term of the sequence.The first term in the sequence 10, 20, 30, 40 is 10, the second term is 20, while thethird term is 30 and the fourth term is 40. It is also good to point out that thepreceding term of a given term is the term immediately before that given term. Forexample, in the given sequence 20 is the term that precedes 30. Examples of other sequences are shown below. These sequences areseparated into two groups. A finite sequence contains a finite number of terms. Aninfinite sequence contains an infinite number of terms.Finite sequence Infinite sequence1, 1, 2, 3, 5, 8 1, 3, 5, 7, …1, 2, 3, 4, 5, …, 8 1, 1 , 1 , 1 , … 2481, -1, 1, -1 1, 1, 2, 3, 5, 8, …In general: A sequence is a set of numbers written in a specific order: a1, a2, a3, a4, a5, a6,………, an The number a1 is called the 1st term, a2 is the 2nd term, and in general, an is thenth term. Note that each term of the sequence is paired with a natural number. Given at least the first 3 terms of a sequence, you can easily find the next termin that sequence by simply discovering a pattern as to how the 3rd term is derivedfrom the 2nd term, and the 2nd from the 1st term. You will find that either a constantnumber is added or subtracted or multiplied or divided to get the next term or acertain series of operations is performed to get the next term. This may seem hard atfirst but with practice and patience in getting them, you will find that it’s very exciting. 3

Examples:Find the next term in each sequence. 1. 17, 22, 27, 32, … 2. 1 , 1 , 1 , 1 … 2 5 8 11 3. 5, 10, 20, 40,… 4. 3, -3, 3, -3,…Solutions: 1. Notice that 5 is added to 17 to get 22, the same is added to 22 to get 27, and the same (5) is added to 27 to get 32. So to get the next term add 5 to the preceding term, that is, 32 + 5 = 37. The next term is 37. 2. Notice that 1 is the numerator of all the fractions in the sequence while the denominators- 2, 5, 8, 11 form a sequence. 3 is added to 2 to get 5, 3 is also added to 5 to get 8. So that 3 is added to 11 to get 14. The next term is therefore 1/14. 3. For this example, 2 is multiplied to 5 to get 10, 2 is multiplied to 10 to get 20 and 2 is also multiplied to 20 to get 40. So the next term is 80, the result of multiplying 40 by 2. 4. It is easy to just say that the next term is 3 since the terms in the sequence is alternately positive and negative 3. Actually the first, second, and third terms were multiplied by -1 to get the second, third and fourth terms respectively.Try this out A. Write F if the sequence is finite or I if the sequence is infinite. 1. 2, 3, 4, 5, ….., 10 2. 7, 10, 13, 16, 19, 22, 25 3. 4, 9, 14, 19, … 4. 2, 6, 18, 54 5. 3, 9, 27, 81, …., 729, … 6. -2, 4, -8, 16, ….. 7. 100, 97, 94, 91, …, -2 8. 1 , 1 , 1 , 1 , 1 4 8 16 32 64 9. 1, 4, 9, 16, 25, …., 144 10. 1 , 2 , 3 , 4 4 9 16 25 4

B. Answer the puzzle. Why are Policemen Strong? Find the next number in the sequences and exchange it for the letter whichcorresponds each sequence with numbers inside the box to decode the answer tothe puzzle.A 2, 5, 11, 23, __ N 2, 6, 18, 54, __B 2, 4, 16, __ O 20, 19, 17, __C 7, 13, 19, __ P 2, 3, 5, 7, 9, 11, 13, 15, __D 19, 16, 13, __ R 13, 26, 39, __E 4, 8, 20, 56, __ S 5, 7, 13, 31, __F 2, 2, 4, 6, 10, 16, __ T 1, 1, 2, 4, 7, 13, 24, __H 1, 1, 2, 4, 7, 13, __ U 1, 1, 1, 2, 3, 4, 6, 9, 13, __I 3, 6, 12, 24, __ Y 1, 2, 2, 4, 3, 6, 4, 8, 5, 10, __L 10, 11, 9, 12, 8, __256 164 25 47 19 85 164 44 24 164 6 25 47 16224 14 13 10 19 17 44 52 47 26 26 48 25 Lesson 2 Finding the Terms of a Sequence Frequently, a sequence has a definite pattern that can be expressed by a ruleor formula. In the simple sequence 2, 4, 6, 8, 10, ….each term is paired with a natural number by the rule an = 2n. Hence the sequencecan be written as 2, 4, 6, 8,… 2n,…1st term 2nd term 3rd term 4th term nth term a1 a2 a3 a4 an 5

Notice how the formula an = 2n gives all the terms of the sequence. Forinstance, substituting 1, 2, 3, and 4 for n gives the 1st four terms: a1 = 2(1) = 2 a3 = 2(3) = 6 a2 = 2(2) = 4 a4 = 2(4) = 8To find the 103rd term of this sequence, use n=103 to get a103 = 2(103) = 206.Examples:1. Find the 1st four terms of the sequence whose general term is given by an = 2n – 1.Solution: To find the first, second, third and fourth terms of this sequence, simplysubstitute 1, 2, 3, 4 for n in the formula an = 2n-1.If the general term is an = 2n – 1, then the 1st term is a1 = 2(1) – 1 = 1 2nd term is a2 = 2(2) – 1 = 3 3rd term is a3 = 2(3) – 1 = 5 4th term is a4 = 2(4) – 1 = 7. The 1st four terms of this sequence are the odd numbers 1, 3, 5, and 7. Thewhole sequence can be written as 1, 3, 5, …, 2n – 1 Since each term in this sequence is larger than the preceding term, we saythat the sequence is an increasing sequence.A sequence is increasing if an + 1 > an for all n.2. Write the 1st 4 terms of the sequence defined by an = 1 . n +1Solution:Replacing n with 1, 2, 3, and 4, respectively the 1st four terms are: 1st term = a1 = 1 = 1 1+1 2 6

2nd term = a2 = 1 = 1 2+1 3 3rd term = a3 = 1 = 1 3+1 4 4th term = a4 = 1 = 1 4+1 5 The sequence defined by an = 1 can be written as n +1 1 , 1 , 1 , 1 , ……, 1 2345 n +1 Since each term in the sequence is smaller than the term preceding it, thesequence is said to be a decreasing sequence. A sequence is decreasing if an + 1 < an for all n.3. Find the 1st 5 terms of the sequence defined by an = (−1) n . 2nSolution: Again by simple substitution, 1st term = a1 = (−1)1 =-1 21 2 2nd term = a2 = (−1) 2 =1 22 4 3rd term = a3 = (−1) 3 = -1 23 8 4th term = a4 = (−1) 4 =1 24 16 5th term = a5 = (−1) 5 =- 1 25 32 The sequence defined by an = (−1) n can be written as 2n - 1 , 1 , - 1 , 1 , - 1 ,…, (−1)n 2 4 8 16 32 2n 7

Notice that the presence of (-1) in the sequence has the effect of makingsuccessive terms alternately negative and positive. It is often useful to picture a sequence by sketching its graph w/ the n valuesas the x- coordinates and the an values as the y- coordinates. The corresponding graphs of Examples 1 - 3 are given below Example 1 8 7 6 5a 4 n 3 2 1 0 0 1 23 4 5 n Example 2 0.6 0.5 0.4 a n 0.3 0.2 0.1 0 012345 n 8

Example 3 0.4 0.2 0 an 0 1 2 3 4 5 6 -0.2 -0.4 -0.6 n You can also find a specific term, given a rule for the sequence, as seen in thefollowing example.4. Find the 13th and 100th terms of the sequence whose general term isgiven by An = (−1) n n2Solution: For the 13th term, replace n with 13 and for the 100th term, replace n w/ 100: 13th term = a13 = (−1)13 = −1 132 169 100th term = a100 = (−1)100 =1 100 2 10000Try this outA. Write the 1st 4 terms of the sequence whose nth term is given by the formula.1. an = n + 1 2. an = 2 – 2n3. an = n - 1 4. an = 2 n5. an = 2n + 1 6. an = n² + 17. an = 3n - 1 8. an = n9. an = 1 - 2n n +111. an = 3n 10. an = n – 1 n 12. an = (-1) n+1 n 9

13. an = n² - 1 14. an = (-1) n 2 n15. an = n² - 1 16. an = n 2 − 1 n n17. an = (−1) n+1 18. an = (−1) n+1 n +1 n2 +119. an = 2  1  n+1 20. an = 1 n³ + 1 3 321. an = (1)n (n²+2n+1)B. Find the indicated term of the sequence whose nth term is given by the formula.22. an = 3n + 4 a12 a1123. an= n(n -1) a1524. an= (-1) n - 1 n² a8 a1025. an= ( 1 )n a12 2 a25 a526. an = 2n - 5 a17 a727. an = n a6 n +1 a1628. an = (-1) n - 1 (n - 1) a829. an = ( 2 ) n 3 a630. an = (n + 2)(n + 3)31. an = (n + 4)(n + 1)32. an = (−1) 2n+1 n233. an = (−1) 2n n+434. an = 3 n² - 2 235. an = 1 n + n² 3 10

Lesson 3 Finding the nth Term of a Sequence In Lesson 2, some terms of a sequence were found after being given thegeneral term. In this lesson, the reverse is done. That is, given some terms of thesequence, try to find the formula for the general term.Examples: 1. Find a formula for the nth term of the sequence 2, 8, 18, 32,… Solution: Solving a problem like this involve some guessing. Looking over the first 4 terms, see that each is twice a perfect square: 2 = 2(1) 8 = 2(4) 18 = 2(9) 32 = 2(16) By writing each sequence with an exponent of 2, the formula for the nth term becomes obvious: a1 = 2 = 2(1)² a2 = 8 = 2(2)² a3 = 18 = 2(3)² a4 = 32 = 2(4)² . . . an = 2(n)² = 2n² The general term of the sequence 2, 8, 18, 32,…. is an = 2n². 2. Find the general term for the sequence 2, 3 , 4 , 5 ,…. 8 27 14 Solution: The first term can be written as 2 . The denominators are all perfect cubes 1 while the numerators are all 1 more than the base of the cubes of the denominators: 11

a1 = 2/1 = 1+1 13 a2 = 3/8 = 2 +1 23 a3 = 4/27 = 3+1 32 a4 = 5/64 = 4 +1 43Observing this pattern, recognize the general term to be an = n +1 n33. Find the nth term of a sequence whose first several terms are given1, 3, 5, 7,...2 468Solution: Notice that the numerators of these fractions are the odd numbers andthe denominators are the even numbers. Even numbers are in the formusually written in the form 2n, and odd numbers are written in the form 2n – 1(an odd number differs form an even number by 1). So, a sequence thathas these numbers for its first four terms is given by an = 2n − 1 . 2n4. Find the nth term of a sequence whose first several terms are given -2, 4, -8, 16, -32,…Solution: These numbers are powers of 2 and they alternate in sign, so asequence that agrees with these terms is given by an = (-1)n 2n.Note: Finding the nth term of a sequence from the 1st few terms is not always automatic. That is, it sometimes takes a while to recognize the pattern. Don’t be afraid to guess the formula for the general term. Many times an incorrect guess leads to the correct formula. 12

Some pointers on how to find the general term of a sequence is given below.Pointers on How to Find the General Term of a Sequence1. Study each term of the sequence as it compares to its term number. Then answer the following questions:a. Is it a multiple of the term number?b. Is it a multiple of the square or cube of the term number? If each term is a multiple of the term number, there will be a common number.2. Examine the sequence. Does it increase or decrease?a. If it increases slowly, consider expressions that involve the term number plus or minus a constant like: n + 2 or n – 3.b. If it increases moderately, think about multiples of the term number plus or minus a constant like: 2n or 3n – 1.c. If the sequence increases very rapidly, try powers of the term number plus or minus a constant like: n2 or n2 + 1.3. If the sequence consists of fractions, examine how the denominator andnumerator change as separate sequences. For example: an = n +1 yields n22 , 3 , 4 , 5 , 6 ,...1 4 9 16 25 Also, though not all sequences can be defined by a formula, like for thesequence of prime numbers, be assured that the sequences discussed or given hereare all obvious sequences that one can find a formula or rule for them.Try this outA. Write the formula for the nth term of the sequence: 1. The sequence of the natural numbers. 2. The sequence of the negative even integers. 3. The sequence of the odd natural numbers. 4. The sequence of the negative odd numbers. 5. The sequence of the multiples of 7. 6. The sequence of the positive even integers that are divisible by 4. 7. The sequence of the negative integers less than -5. 13

8. The sequence of the positive odd integers greater than 9.9. The sequence for which a1 is 8 and each term is 6 more than the preceding term.10. The sequence for which a1 is 3 and each term is 10 less than the preceding term.B. Determine the general term for each of the following sequences:1. 2, 3, 4, 5,… 12. 3, 9, 27, 81,…2. 3, 6, 9, 12,… 13. -2, 4, -8, 16,…3. 4, 8, 12, 16, 20,… 14. -3, 9, -27, 81,…4. 3, 4, 5, 6,… 15. 1 , 1 , 1 , 1 ,…5. 7, 10, 13, 16,…6. 4, 9, 14, 19,… 4 8 16 327. 3, 12, 27, 48,… 16. 1 , 1 , 1 , 1 ,…8. 2, 16, 54, 128,…9. 4, 8, 16, 32,… 3 9 27 8110. 1, 8,27, 64,… 17. 1 , 2 , 3 , 4 ,…11. 1, 4, 9, 16,… 4 9 16 25 18. 1 , 2 , 3 , 1 ,… 4 10 28 32 Lesson 4Recursively Defined Sequences Some sequences do not have simple defining formulas like those in Lesson 3.The nth term of a sequence may depend on some or all of the terms preceding it. Asequence defined in this way is called recursive. One has to be sure that the notations: an and an – 1 is understood first beforegoing to the examples below. To illustrate: if the 24th term is an, then it can be writtenas a24. So that an – 1 means a24 – 1 = a23 or the 23rd term.Examples: 1. Find the first five terms of the sequence defined recursively by a1 = 1 and an = 3(an – 1 + 2). Solution: The defining formula asks that the preceding term, an – 1, be identified first before one can get the nth term, an. Thus, one can find the second term from the 14

first term, the third term from the second term, the fourth term from the third term, and so on. Since the first term is given as a1 = 1, proceed as follows: a2 = 3(a1 + 2) = 3(1 + 2) = 9 a3 = 3(a2 + 2) = 3(9 + 2) = 33 a4 = 3(a3 + 2) = 3(33 + 2) = 105 a5 = 3(a4 + 2) = 3(105 + 2) = 321 Thus, the first five terms of this sequence are 1, 9, 33, 105, 321,… 2. Find the first 11 terms of the sequence defined recursively by F1 = 1, F2 = 1 and Fn = Fn – 1 + Fn – 2 Solution: To find Fn, the preceding terms, Fn – 1 and Fn – 2 have to be found first. Since the first two terms are given, then F3 = F2 + F1 = 1 + 1 = 2 F4 = F3 + F2 = 2 + 1 = 3 F5 = F4 + F3 = 3 + 2 = 5 By this time it should be clear as to what is happening here. Each term issimply the sum of the first two terms that precede it, so that its easy to write down asmany terms as one pleases. Here are the first 11 terms: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,… The sequence in Example 2 is called the Fibonacci sequence, named afterthe 13th century Italian mathematician who used it to solve a problem about thebreeding of rabbits. The sequence also occurs in a lot of other applications in nature.The Fibonacci sequence 8In the branching of a tree 5 3 2 1 1 15

To better understand the difference between recursively defined sequencesand the general sequences encountered in the last three lessons, study theexamples below. 3. Find the 120th term of the sequence defined by an = 2n + 3.Solution:Replacing n with 120, then 120th term = a120 = 2(120) + 3 = 243 The 120th term is 243. Notice that the 120th term from the example above is found without finding firstthe preceding term, that is, the 119th term! Whereas with the recursively definedsequence one has to find all the terms preceding the required term,4. Find the 4th term of the recursively defined sequence an = an−1 where a1 = -3. 2Solution: First term = a1 = -3 Second term = a2 = a2−1 = a1 = − 3 2 22 −3 Third term = a3 = a3−1 = a2 = 2 = − 3 22 24 −3 Fourth term = a4 = a4−1 = a3 = 4 = − 3 2 22 8The fourth term is - 3 . 8Try this outFind the first five terms of the given recursively defined sequences: 1. an = 2(an – 1 – 2) and a1 = 3 and a1 = -8 2. an = an−1 and a1 = 1 2 16 3. an = 2an – 1 + 1

4. an = 1 and a1 = 1 1 + an−1 and a1 = 1, a2 = 25. an = an – 1 + an – 2 and a1 = a2 = a3 = 16. an = an – 1 + an – 2 + an – 3 Lesson 5Arithmetic Sequences There are two main types of sequences. These are the arithmetic sequencesand the geometric sequences. This lesson will show what arithmetic sequences are.The more detailed lesson on arithmetic sequences will be discussed in the nextmodule. Look at the following sequences. 1. 4, 7, 10, 13,… 2. 33, 38, 43, 48,… 3. -2, -6, -10, -14,… 4. 100, 98, 96, 94,… 5. 1 , 1, 1 1 , 2, … 22 Can you give the next two terms of the above sequences? How did you getthe next terms? If you get 16 and 19 for a, then you are correct. Notice that a constant number, 3, is added to each term to get the next term. In b, 5 is added to the preceding termafter the first, while in c, -4 is added to get the next term, in d, -2 is added to thepreceding term and in e, ½ is added to get the next term. Notice that a constant or common number is added to the preceding term toget the next term in each of the sequences above. All these sequences are calledarithmetic sequences. The constant number is called the common difference and isrepresented as d. To find the common difference, d, simply subtract the first term from thesecond term, a2 – a1, or the second term from the third term, a3 – a2, or the thirdterm from the fourth term, a4 – a3; or in general, d = an – an – 117

Examples:1. Determine if the sequence is arithmetic or not. If it is, find the common difference and the next three terms. Then graph. -11, -4, 3, 10,…Solution: To find out if the sequence is arithmetic, there must be a commondifference between any two terms in the sequence. So that d = a2 – a1 = -4 – (-11) = 7 = a3 – a2 = 3 – (-4) = 7 = a4 – a3 = 10 – 3 = 7 The sequence is arithmetic and the common difference is 7. The nextthree terms are obtained by adding 7 to the preceding term, so that a5 = a4 + 7 = 10 + 7 = 17 a6 = a5 + 7 = 17 + 7 = 24 a7 = a6 + 7 = 24 + 7 = 31 Example 1 40 30 20 an 10 0 -10 0 2 4 6 8 -20 n2. Write the first five terms of the arithmetic sequence with first term 5 and common difference -2.Solution: The second term is found by adding -2 to the first term 5, getting 3. For thenext term, add -2 to 3, and so on. The first five terms are 5, 3, 1, -1, -3. 18

Remark: There is another way of finding the specified term of an arithmetic sequencebut it will be discussed in the next module. The same thing is true for the generalterm of any arithmetic sequence.Try this out Determine whether the sequence is arithmetic or not. If it is, find the commondifference and the next three terms. 1. 2, 5, 8, 11,… 2. 2, -4, 6, -8, 10,… 3. -6, -10, -14, -18,… 4. 40, 42, 44, 46,… 5. 1.2, 1.8, 2.4,… 6. 1, 5, 9, 13,… 7. 1 , 1 , 1 , 1 ,... 2345 8. 5, 6, 7, 8,... 9. 98, 95, 92, 89,… 10. 1, 4 , 5 , 2,… 33 Let’s Summarize 1. A sequence is a list of numbers in which order is important. a1, a2, a3, a4, …, an, … Each number in the list corresponds to each natural number. 2. A sequence may either be finite or infinite. A finite sequence has a specific number of terms. An infinite sequence has an endless number of terms. 3. To find the terms of a sequence given its rule, simply replace n with the number of the specific term needed to be found. 4. Most sequences have a general term or rule that describes all the terms in the sequence. There is no specific way of finding the general term of a given sequence. 5. A sequence is defined recursively when the nth term can be found only when the preceding term is found. 19

6. An arithmetic sequence is a sequence where each term is found by adding a constant number, called the common difference, to the preceding term.7. The Fibonacci sequence is a special recursively defined sequence.What have you learnedA. Write the first five terms of the sequence.1. an = n + 7 3. an = 5n – 2 n2. an = 3n - 1B. Find the indicated term for the sequence.4. an = -7n + 3; a8 6. an = 2n + 7 ; a145. an = (n + 2)(2n - 3); a5 3n − 5C. Find the general term, an, for the given terms of the sequence.7. 3, 7, 11, 15,… 9. 1 , 1 , 1 , 1 ,...8. 0, -4, -8, -12,… 2 4 8 16E. Find the first four terms of the sequence defined recursively.10. a1= -1, an = 5an – 1 11. a1 = 6, an = an−1 3n − 1F. Find the common difference and the next three terms of the given arithmetic sequence.12. 1, 10, 19, 28,… 14. 1, 3, 5, 7,…13. 5.5, 7, 8.5, 10,… 15. 43, 39, 35, … 20

Answer KeyHow much do you know 9. an = 1 n2 + n 1 4, 5 , 2 , 7 , 8 2 45 10 -1, -3, -9, -27 2 1, 2, 4, 8, 16 11 5, 5, 5 , 5 3 1, 5, 9, 13, 17 26 4 -70 12 d = 6; 26, 32, 38 5 78 6 49/23 13 d = 1.5; 13, 14.5, 16 7 an = 4n 8 an = -10n 14 d = 1; 5, 6, 7Lesson 1 15 d = -3; 15, 12, 9 A.1F 3I 5I 7 F 9F2F 4F 6I 8 F 10 FB. Because they can hold up traffic.Lesson 2A.1 2, 3, 4, 5 12 1, -2, 3, -42 0, -2, -4, -63 0, 1, 2, 3 13 0, 3, 8, 154 2, 4, 8, 16 14 -2, 4, -8, 165 3, 5, 7, 9 15 0, 7 26 636 2, 5, 10, 17 ,, 23 47 2, 5, 8, 11 16 0, 3 8 15 ,, 23 4 17 1 ,− 1 , 1 ,− 1 2 34 5 18 1 ,− 1 , 1 ,− 1 2 5 10 17 21

8 1,2,3,4 19 2 , 2 , 2 , 2 2345 9 27 81 243 9 -1, -3, -5, -7 20 4 ,11,10, 67 33 3 10 0, 3 , 8 ,15 23 4 21 4, 9, 16, 25 11 3, 9, 27, 81 29 32 243 B. 30 380 22 40 31 88 32 −1 23 110 24 225 36 25 1 33 1 256 20 26 15 34 94 27 12 35 38 13 6 an = 4n 28 24 7 an = -(n + 5) 8 an = 2n + 9Lesson 3 9 an = 6n + 2 10 an = -10(n + 1)+ 3 A. 10 an = n3 1 an = n 11 2 an = -2n 12 an = n2 3 an = 2n – 1 13 4 an = -2n + 1 14 an = 3n 5 an = 7n 15 an = (-2)n B. 16 an = (-3)n 1 an = n + 1 22 an = 1 2 an = 3n 2 n+1 3 an = 4n 4 an = n + 2 1 5 an = 3n + 4 an = 6 an = 5n – 1 3n 7 an = 3n2

8 an = 2n3 17 an = n (n + 1)2 9 an = 2n + 1 18 an = nLesson 4 3n +1 1 3, 2, 0, -4, -12 4 1, 1 , 2 , 3 , 5 2358 2 -8, -4, -2, -1, - 1 2 5 1, 2, 3, 5, 8 3 1, 3, 7, 15, 31 6 1, 1, 1, 3, 5Lesson 5 d=3 14, 17, 20 1 Arithmetic d = -4 -22, -26, -30 2 No d=2 48, 50, 52 3 Arithmetic d = 0.6 3, 3.6, 4.2 4 Arithmetic d=4 17, 21, 25 5 Arithmetic 6 Arithmetic d=3 86, 83, 80 7 No d = 1/3 7, 8,3 8 No 33 9 Arithmetic 10 Arithmetic 9. an = 1 2nWhat have you learned 10 -1, -5, -25, -125 1 8, 9 ,10 ,11,12 23 4 5 11 6, 63 3 ,, 2 1, 3, 9, 27, 81 5 20 220 3 3, 8, 13, 18, 23 12 d = 9; 37, 46, 55 4 -53 5 49 13 d = 1.5; 11.5, 13, 14.5 6 35/37 7 an = 4n – 1 14 d = 2; 9, 11, 13 8 an = 4(1 – n) 15 d = -4; 31, 27, 23 23

Module 1 Variations What this module is about Quantities are related to each other in some ways. Some quantities areconstants and others are variables. This module will describe these relatedquantities. Two quantities are related if a change in the value of one correspondsto a predictable change in the value of the other. The direct and direct squarevariations are the focus of this module. What are you expected to learn 1. Determine the relation between two quantities. 2. Identify quantities that are related directly to each other. 3. Find the equation of the variation and its constant of variation. 4. Use proportion to solve direct variation. How much do you knowA. Which of the following situations show direct variation. Write Y if it is and N if not. 1. The diameter of a circle is related to its circumference. 2. Water pressure on a submarine depends on the depth. 3. The number of people sharing a pizza is related to the size of the slices of the pizza. 4. The area of the wall is related to the amount of paint used to cover it. 5. The area of the face of a cube is related to its volume. 6. The time a teacher spends checking papers is related to the number of students. 7. The cost of life insurance depends on the age of the insured person.

8. The age of a used car is related to its resale value.9. The amount of money raised for a fundraiser is related to the number of tickets sold.10. The distance an airplane fly depends on its time of travel.B. Express the following ratios in lowest terms.1. 10 :15 5. 2 : 82. 0.3 : 6 6. 1000 to 100003. 18 7. 250 : 500 8. 120 :140 12 9. 0.04 : 0.084. 2x 10. 0.005 : 0.010 3x What will you do Lesson 1 Direct Variation A bicycle is traveling 10 kilometers per hour (kph). In one hour it goes 10kilometers (km). In two hours, it goes 20 km. In 3 hours, it goes 30 km and so on.Using the number of hours as the first number and the number of kilometerstraveled as the second number: (1,10), (2,20), (3,30), (4,40) and so on. Note thatas the first number gets larger, so does the second. Note also, that the ratio ofdistance to time for each of these ordered pairs is a constant, 10 or 10. 1 There is direct variation whenever a situation produces pairs of numbersin which their ratio is constant. Hence, the distance traveled by the bicycle variesdirectly as the time of travel. d = 10 (10 is a constant) or d = 10t t 2

The statements: “y varies directly as x.” “y is directly proportional to x.” and “y is proportional to x.” Imply direct variation and are used in many situations. Likewise thestatements translate mathematically as y = kx . Where k is often referred to asthe constant of proportionality. When there is direct variation the variation constant can be found if onepair of values of x and y is known. After that other pairs of values can also befound.Examples: 1. If y varies directly as x and that y = 32 when x = 4 . Find the variation constant and the equation of variation. Solution: Express the statement “y varies directly as x”, as y = kx . Substituting the given values in the equation, you have: 32 = k(4) k = 32 4 k =8The variation constant is 8 and the equation of variation is given by: y = 8x2. Find the equation of variation when y varies directly as x and y = 7 whenx = 25 .Solution: y = kx 7 = k(25) Substitute the given. 3

k = 7 or 25 k = 0.28Therefore the equation of the variation is y = 7 x or y = 0.28x 253. The circumference of a circle varies directly with its diameter. If the circumference of 7-diameter circle is 7π , what is the circumference of the circle whose diameter is 10cm? 15 cm? 18cm? 20cm?Solution1: The statement “ The circumference of a circle varies directly with itsdiameter.” can be expressed as c = kd . Substituting the given values of cand d in the equation: 7π = k(7) k = 7π 7 k =π Therefore, c = kd can be expressed as c = π d by substituting π inplace of k. To solve for c: when d = 10 cm, c = π (10) c = 10π cm when d = 15 cm, c = π (15) c = 15π cm when d = 18 cm, c = π (18) c = 18π cm 4

when d = 20 cm, c = π (20) c = 20π cmSolution 2: Since their ratio is always constant, c = k , you can derive the dfollowing proportions: c1 = c2 ; c1 = c3 ; c1 = c4 ; c1 = c5 d1 d2 d1 d3 d1 d4 d1 d5 In tabular form, the given values can be expressed as: d 7 10 15 18 20 c 7π c2 c3 c4 c5 Let d1 = 7 , c1 = 7π , d2 = 10 , d3 = 15 , d4 = 18 , and d5 = 20 . You need to solve for c1 = c2 . Substituting the given values in the d1 d2proportion, you have 7π = c2 7 10 π = c2 the 7’s cancel out 10 multiply each term by the LCD π = c2  10 to eliminate the denominators 10  c2 = 10π cmFor c3 , use the proportion c1 = c3 . Substitute the given values in the d1 d3proportion: 7π = c3 7 15 5

π = c3 the 7’s cancel out 15 multiply each term by the LCD π = c3  15 to eliminate the denominators 15  c3 = 15π cmFor c4 , use the proportion c1 = c4 . Substitute the given values in the d1 d4proportion: 7π = c4 7 18 π = c4 the 7’s cancel out 18 π = c4  18 multiply each term by the LCD 18  to eliminate the denominators c4 = 18π cm Therefore, if the circumference of a circle varies directly as its diameter,then when the diameter of the circle is 7 cm, 10 cm, 15 cm, 18 cm and 20 cm,the corresponding circumference is 7π cm, 10π cm, 15π cm, 18π cm and20π cm.4. y varies directly as x , and x = 12 when y = 5 . What is the value of y when x = 20 ?Solution 1: “ y varies directly as x ” is expressed as y = kx . There are twovariables; x and y. Knowing one set of values for x and y, you can find k .Using this value for k and the other value given for x , you can find thecorresponding value for y .Substituting the first set of values in y = kx , you have: 5 = k(12) k= 5 12 6

The equation of the variation is y = 5 x 12Solving for y when x = 20 : y = 5 (20) 12 y = 5 (20)5 3(12) y = 25 or 3 y =81 3Solution 2:From example 3, you can use a similar proportion: y1 = y2 x1 x2Substituting the values, you get 5 = y2 12 20 5 = y2  60 multiply each term by the LCD to 12 20  eliminate the denominators (60)5 = (60) y2 12 20 5(60)5 = 3(60) y2 12 20 25 = 3y2 y2 = 25 or 3 y =81 35. Find the constant of variation and the equation of variation. If y varies directly as x , and y = 15 when x = 3 . 7

Solution: The problem states that y = kx . Substitute the values to solve fork , you have 15 = k(3) k = 15 3 k =5 Hence, the equation of the variation: y = 5x6. x varies directly as y . If x = 35 when y = 7 , what is the value of y when x = 25 ?Solution 1. The problem states that x = ky , substitute the first values of y and xrespectively to get k , you have: 35 = k(7) k = 35 7 k =5 Hence the equation of the variation: x = 5y Solving for y when x = 25 , 25 = 5y y = 25 5 y=5Solution 2: From example 4, you can use a similar proportion: y1 = y2 x1 x2Where x1 = 35 , y1 = 7 and x2 = 25 8

Substituting the values, you get 7 = y2 35 25 7 = y2 35 25 1 = y2 7 reduced to lowest term is 1 5 25 35 5 y2 = 25 5 y2 = 5Try this outA. An ordered pair in a direct variation is given; such that the first varies directlyas the second term in the number pair below. State the constant ofproportionality.1. (4,28) 6. (12, - 15)2. (5,9) 7. (1/3, 1)3. (1, 1/3) 8. (3d, d)4. (c, 2c) 9. (6, 27)5. (15,3) 10. (36, 24)B. Find an equation where y varies directly as x.1. y = 28 when x = 7 7. y = 200 when x = 3002. y = 30 when x = 8 8. y = 1 and x = 23. y = 0.7 when x = 0.4 9. y = 48 and x = 64. y = 0.8 when x = 0.5 10. y = 10 and x = 245. y = 400 when x = 1256. y = 630 when x = 175 9

C. Solve for each of the following. Use any method. y varies directly as x. find thevalues as indicated.1. If y = 12 when x = 4, find y when x = 122. If y = -81 when x = 9, find y when x = 73. If y = −3 when x = −4 , find x when y = 24. If y = 3 when x = 10, find x when y = 1.25. If y = 2.5 when x = 7.25, find y when x = 4.75 Lesson 2Direct Square Variation The number of ways in which one variable depend upon another isunlimited. One of which is when one quantity varies directly as the square ofanother. That is y varies directly as the square of x. The following examples are presented for you to gain skill in solving suchtype of variation.Examples: 1. y varies directly as the square of x , and y = 10 , when x = 4 , find the constant and equation of the variation. Solution:To express the statement “ y varies directly as square of x ”, write:y = kx2 ,Substitute the given values in the equation, you have:10 = k(4)210 = 16kk = 10 1610

k=5 8 The variation constant is 8 and the equation of variation is given by 5 y = 8 x2 52. y varies directly as the square of x . Find the constant and equation of variation if y = 12 when x = 5 .Solution: Substitute the given values in the equation, you have: 12 = k(5)2 12 = 25k k = 12 25 Hence, the equation of the variation y = 12 x2 253. y varies directly as the square of x . If y = 3 when x = 2 , find y when x=7.Solution: The equation suggests y = kx2 . Substitute the first set of values in y = kx2 3 = k(2)2 3 = 4k k=3 4 The equation of the variation is y = 3 x2 4 11

Solving for y when x = 7 : y = 3 (7)2 4 y = 3 (49) 4 y = 147 or 4 y = 36 3 4To check, both equations of variation should have the same value of k . y =3, x = 2 y = 147 , x = 7 k= y 4 x2 k= y x2 147 k = 3 k = 4 (2)2 (7)2 k=3 147 4 k= 4 49 k = 147 1 4 49 k=3 44. y varies directly as the square of x . If y = 9 when x = 6 , find x when y = 4.Solution:“ y varies directly as the square of x ” translates to: y = kx2 .From the given values: 9 = k(6)2 9 = 36k k= 9 36 12

k=1 4The equation of the variation: y = 1 x2 4Looking for x in the second set of values: 4 = 1 x2 4 4 = 1 x2  4 Apply the multiplication property of 4  equality16 = x2 Take only the principal (positive) root. x = ± 16 x = ±45. y varies directly as the square of x . If y = 1 when x = 8 , find x when y = 4.Solution: From the given values: 1 = k(8)2 1 = 64k k= 1 64The equation of the variation: y = 1 x2 64Solving for x in the second set, substitute the values: y = 1 x2 64 4 = 1 x2 64 x2 = 464 13

x2 = 256 x = 256 x = 16 To check if both set of values have the same constant of variation k ,substitute the values in x and y in the equation, y = kx2 .1st set 2nd set1 = k(8)2 4 = k(16)21 = 64k 4 = 256kk= 1 k= 4 64 256 k= 1 64 In general, direct square variation can be extended to any n th power of a variable provided that n > 0 . The statement, “ y varies directly as the nth power of x ,” translates as y = kxn .Try this outA. Write an equation to describe these situations. 1. y varies directly as x . 2. A varies directly as the square of p . 3. V varies directly as the cube of x . 4. T varies directly as the square root of a . 5. N varies directly as s . 6. R varies directly as the fourth root of M 7. The surface area A of a sphere varies directly as the square of the radius r. 14

B. What kind of boats do vampires like?Answer the questions below then transfer the letter associated to each questionto the box, which contains the correct answer to decode the key.Y varies directly as the Find the variation constant Y varies directly as x.square of x. Find k when if y = 48 and x = 6 if y Find y if x = 3, whenx = 3 and y = 12. varies directly as the y = 5 if x = 2. square of x.S B EFind an equation of Write an equation for thevariation if y varies Find y1 in the proportion: phrase: The electricdirectly as x and y = 15 current I in an electricwhen x = 4. y1 = 12 circuit varies directly as 45 the Voltage V .S L LThe perimeter of asquare varies directly as If the constant of a direct Write the equation ofits side. Find the variation is 15 , find x variation: The distance (d) inconstant of variation. km covered by a train varies 4 directly as its speed (s) inS when y = 10. kph. O DIf the constant of If y varies directly as x Determine the constant ofvariation 5, find y when and y = 10 when x = 5, variation in the table:x = 7. find y when x = 15. x2 4 68EO y 5 10 15 20 If y varies directly as x V15 48 30 2 2 d=ks 2 4 y = 15 x k= p 35 I = kV 42 5 35 3 4 s 3 15

C. Solve the following: 1. A varies directly as the square of b and A = 8 , when b = 2 . Find A when b=7. 2. The circumference of a circle C varies directly as the radius r of the circle. The circumference is 10π cm, when the radius if 5 cm. Write an equation and determine the constant of variation. 3. x varies directly as y3 . If x = 64 when y = 2 , find x when y = 1.Let ‘s summarizeDefinition: A variable y varies directly as variable x if and only if y = kx . Theconstant k is called the proportionality constant. In the same manner, that a variable y varies directly as the squareof x is expressed as y = kx2 .Direct variation is also related to proportion. Let (x1,y1) and (x2, y2)be any two solutions for y = kx , then y1 = kx1 or y1 = k . Similarly you x1have y2 = kx2 or y2 = k . Since we can equate y1 and y2 to k , it follows x2 x1 x2that y1 = y2 . x1 x2What have you learnedA. Multiple Choice: Choose the letter of the best answer.1. If p varies directly as q , the equation can be written asa. pq = k b. k = p c. p = k d. pk = q q q 16

2. Which equation shows that p varies directly as q ?a. p = kq b. pq = k c. q = k d. q = pk p3. If m and n are dependent and independent variables respectively, which equation shows direct variation.a. mn = k b. m = kn c. m = k d. n = mk n4.The proportion y1 = y2 is an equation of ____________variation. x1 x2a. direct b. inverse c. joint d. square5. In the equation A = π r2, the constant of variation isa. A b. r c. π d. 2B. Determine if the table expresses a direct variation between the variable. If so, find the constant of variation and an equation that defines the relation.6. x -3 -9 3 6 y1 3 6 -27. x 7 14 -21 -28 y3 6 -9 -128. A -15 10 -20 25 R -3 2 -4 59. R 2 3 4 5 S1 2 3 410. x 16 20 24 28 32 y 12 15 18 21 24C. Solve for the following.11. If b is directly proportional to x and b = 8 when x = 3, find b when x = 12.12. If y varies as x2 and if y = 40, when x = 2, find the value of x when y = 10. 17

Answer key 6. Y 7. YHow much do you know 8. N – usually true but not for 1. Y vintage/classic cars 2. Y 9. Y 3. N 10. Y 4. Y 6. 1 to 10 5. Y 7. 1: 2 8. 6 : 7B. 9. 1: 2 1. 2 : 3 10. 1: 2 2. 1: 20 3. 3 6. − 5 2 4 4. 2 3 7. 3 5. 1: 4 8. 1Try this out 3 9. 9Lesson 1 2A. 1. 7 10. 2 2. 9 5 3 3. 1 6. y = 15 x 3 4. 2 4 5. 1 7. y = 15 x 5 4B. 1. y = 4x 8. y = 15 x 2. y = 15 x 4 4 18 3. y = 15 x 4

4. y = 15 x 9. y = 8x 4 10. y = 15 x 5. y = 15 x 4 4C. 1. y = 36 2. y = −63 3. x = 8 3 4. x = 4 5. y = 95 58Lesson 2A. 1. y = kx 2. A= kp2 3. V= kx3 4. T = k a 5. N= ks 6. R = k 4 m 7. A= kr2B. What kind of boats do Vampires like?B L O O D VE S SE L S15 48 30 2 2 d=ks 2 4 y = 15 x k= p 35 I = kV 42 5 35 3 4 s 3C. 1. A = 98 2. C = 2Πr 3. x = 8 19

What have you learnedA. 1. b 2. a 3. b 4. a 5. cB. 6. no 7. direct variation, k = 3 , y = 3 x 77 8. direct variation, k = 1 , R = 1 x 559. no10. direct variation, k = 3 , y = 3 x 44C. 11. b1 = b2 x1 x28 = b23 12b2 = 8(12) 3b2 = 3212. x = 1 20

Module 1 Radical Expressions What this module is about This module is about radical expressions. This module will require you to recallyour knowledge on exponents and how to relate it to radicals. You will start with thesimple basic lessons on squares and cubes. What you are expected to learn This module is designed for you to demonstrate knowledge and skill insimplifying, performing operations and solving equations and problems involving radicalexpressions. 1. Identify expressions which are perfect squares or perfect cubes. 2. Find the square root or cube root of expressions 3. Given a number in the form m x where x is not a perfect nth root, name two rational numbers between which it lies.How much do you knowA. Find the indicated square roots of the following.1. 169 6. 0.0000812. 400 7. 0.01963. 1225 8. 1.694. 2.25 9. 645. 0.0004 121 10. 441 144

B. Find the cube roots of the following:1. 3 27 6. 64 3 272. 3 125 7. 125 3 83. 3 343 8. 3 − 1 644. 3 0.001 9. 343 3 7295. 3 0.064 10. 3 216 1000C. Name the two rational numbers between which it lies.1. 7 6. 3 652. 15 7. 3 1283. 24 8. 3 2754. 35 9. 3 3805. 48 10. 3 796 What you will do Radical Expressions Recall that the expression bn = x, b is a number multiplied n times to get theproduct x. Some simple situations are taking the squares and cubes of numbers. Takefor example the following:a. Squares of numbers: (4)2 = (4)(4) = 16 x = 4, b = 16, n = 2 (-4)2 = (-4)(-4) = 16 x = -4, b = 16, n = 2 2

b. Cubes of numbers: (2)3 = (2)(2)(2) = 8 x = 2, b = 8, n = 3 (-2)3 = (-2)(-2)(-2) = -8 x = -2, b = 8, n = 3 If we reverse the situation and find the value of b instead or the numbermultiplied n times to get x. This process shall be introduced to you in this section calledradical expressions. Using radical signs, such as are called radical expressions. Some simpleradical expressions are square roots and cube roots. We shall use the radicalsymbol for square roots and 3 for cube roots. This will give you a clearerunderstanding of perfect squares and perfect cubes. Continue reading the lessons. Lesson 1 Identifying Expressions which are Perfect Squares or Perfect CubesPerfect Squares: A number is a perfect square if it is a product of a single number multiplied twotimes by itself.Examples:1. 16 is a perfect square because:(4)(4) or (4)2 = 16 (-4)(-4) or (-4)2 = 162. 64 is a perfect square because: 121 8  8  or  8 2 = 64  − 8  − 8  or  − 8 2 = 641111 11 121  11 11  11 1213. 0.25 is a perfect square because:(0.5)(0.5) or (0.5)2 = 0.25 (-0.5)(-0.5) or (-0.5)2 = 0.25 3

Perfect Cubes: A number is a perfect cube if it is a product of a single number multiplied threetimes by itself.Examples: a. 33 or (3)(3)(3) = 27 b. (-5)3 or (-5)(-5)(-5) = -125c.  2 3 or  2   2   2  =  8   3   3   3   3   27 Try this outA. Identify if the following radical expressions are perfect squares.1. 144 6. -1252. 243 7. 13. 225 84. 1296 8. 5765. 784 529 9. 0.0008 10. 0.025B. Identify if the following radical expressions are perfect cubes.1. 81 6. 3432. -125 7. 5003. 216 8. 0.001 10004. 0.0645. 1000 9. 1345 10. 729C. Tell whether the following radical expressions are perfect squares or perfect cubes or neither.1. 3.24 6. 6.42. 0.1849 7. 0.0013. 7.84 8. 0.5134. 44.1 9. 7595. 15.21 10. 850 4

Lesson 2 Finding the Square Root or Cube Root of Radical ExpressionsSquare Roots: Every positive number has a real number square roots. Negative numbers do nothave real number square roots. The number zero has just one square root and that is 0itself.Consider the expression x : a. x is a rational number, if x is a perfect square. b. x is an irrational number, if x is not a perfect square. The radical expression − x do not have real square roots.Examples:1. 9 is a rational number because 9 is a perfect square.2. 8 is an irrational number because 8 is not a perfect square.3. − 64 is not a real number because no number squared will give you -64. You should be careful not to confuse the idea of the negative square root of anumber with the idea of a square root of a negative number. - 9 = -3 - 9 can also be expressed as – ( 9 ). - 25 = -5 - 25 can be expressed as – ( 25 ). Now let us find square roots of numbers and identify those that do not qualify thegiven conditions.Examples:1. 64 = 8 8 is a square root of 64 since (8)(8) = 64. 64 = -8 -8 is a square root of 64 since (-8)(-8) = 64 5