MATH 2 part 2

Note that the words sequence and progressions are used arbitrarily. Sometimesthis module will use geometric sequence, at other times it will use geometricprogression. In either ways both will mean the same sequence. Definition of a Geometric Progression A sequence an is called a Geometric Progression if there is a nonzero number r such that an = r · an-1 , n > 2 The number r is called the common ratio. Notice that it is defined in terms of the preceding term, an-1. Therefore thesequence is defined recursively. Here are some examples of a geometric progression: 1. 1, 2, 4, 8,… 2. 9, -27, 81, -343,… 3. 1 , 1 , 1 , 1 , 1 2 4 8 16 32 4. .1, .05. 025, .0125 Each is called a geometric progression since there is a common multiplier orcommon ratio, r, between the terms of the sequence after the 1st term. In (a), thecommon ratio, r, is 2; in (b), r = -3; in (c), r = 1 ; and in (d), r = 0.5. 2 The common ratio can be found by dividing any term by its preceding term, thatis, r = a2 = a3 = a4 = a5 = ..... = an a1 a2 a3 a4 a n −1 The sequence 2, 6, 18, 54, 162,… is a geometric progression in which thefirst term, a1, is 2 and the common ratio is r = a2 = a3 = a4 a1 a2 a3 3

r = 6 = 18 = 54 = 3 2 6 18Example 1. Find r and the next three terms of the geometric progression 15, 15 , 15 , 15 ,… 248Solution: To find r, choose any two consecutive terms and divide the second by the first. Choosing the second and third terms of the sequence, r = a3 = 15 ÷ 15 = 15 · 2 = 1 a2 4 2 4 15 2 To find the next three terms, multiply each successive term by 1 2 an = r · an-1 a5 = 1 · a5 - 1 = 1 · a4 = 1 ⋅ 15 = 15 2 2 2 8 16 a6 = 1 · a6 - 1 = 1 · a5 = 1 ⋅ 15 = 15 2 2 2 6 32 a7 = 1 · a7 - 1 = 1 · a6 = 1 ⋅ 15 = 15 2 2 2 32 64 The common ratio is 1 and the next three terms are 15 , 15 , 15 . 2 16 32 64Example 2. Find the first five terms of a geometric sequence whose 1st term is 2 and whose common ratio is -3.Solution: Since a1 = 2 and r = -3, then a1 = 2 a2 = r · a2 - 1 = r◌ּ a1R R = -3 · 2 = -6 a3 = r · a3 - 1 = r· a2 = -3 (-6) = 18 a4 = r · a4 - 1 = r· a3 = -3 (18) = -54 a5 = r · a5 - 1 = r· a4 = -3 (-54) = 162 The 1st 5 terms of the sequence are 2, -16, 18, -54 and 162. 4

Try this out A. Tell whether the following sequences is geometric or not. If geometric, find r. 1. 4, 8, 16, 32,… 2. 1 , 2 , 3 , 4 , 5 ,… 3 33 3 3 3. 1, -3, 9, -27, 81,… 4. 1, − 1 , − 1 , - − 1 , − 1 ,… 2 4 8 16 5. 5, 15, 45, 135,… 6. 1 , 2 , 4 , 8 ,… 3 333 7. 1, -3, 7, -11,… 8. 2 , 2 , 2 , 2 ,… 3 15 75 375 9. 1, -4, 16, -64,… 10. 10, 20, 30, … B. Write the first five terms of the geometric progression where 1. a1 = 2, r = 3 2. a1 = 3, r = 2 3. a1 = 10, r = 1 2 4. a1 = 32, r = 1 4 5. a1 = 3, r = -2 6. a1 = 2, r = -3 7. a1 = 2 , r = - 1 32 8. a1 = 1, r = 0.5 9. a1 = -1, r = 0.5 10. a1 = -2, r = -2 5

Lesson 2 The nth term of a Geometric Progression At this point, a question that could arise is, what if you are asked to find, the 15thterm of a geometric sequence? Does it mean that you have to find first the 14th term toget the 15th term? But, since the 14th term is not yet given, you have to compute the13thterm to get the 14th term. In other words, you have to get first all the terms preceding the15th term. Isn’t there a shorter way of doing this? Actually, there is! There is a rule orformula for the nth term of any geometric sequence. Consider the geometric progression defined by a1 = 2 and r = 3. The 1st fiveterms will be a1 = 2 a2 = r · a1 = 3 · 2 = 6 a3 = r · a2 = 3 · 6 = 18 a4 = r · a3 = 3 · 18 = 54 a5 = r · a4 = 3 · 54 = 162By rewriting this sequence, another pattern is found, as follows:(1) Term Factored Form Exponential Forma1 = (2) (3) (4)a2 = 2a3 = 6 =2 = 2 ·30 *a4 = 18 = 2·3 = 2 ·31a5 = 54 = 2·3·3 = 2 ·32 = 2·3·3·3 = 2 ·33 162 = 2·3·3·3·3 = 2 ·34 * 2 ·30 = 2 ·1 = 2 Notice that column (4) consists of two factors, 2 and 3. These factors are actuallythe first term and the common ratio, respectively, of the given geometric sequence. Witha1 representing the 1st term and r representing the common ratio, column (4) may nowbe written: 6

2 ·30 a1 ⋅ r 0 a12 ·31 a1 ⋅ r 1 a22 ·32 a1 ⋅ r 2 a32 ·33 a1 ⋅ r 3 a42 ·34 a1 ⋅ r 4 a5 Comparing the second column from the third column, notice that the exponent ofthe common ratio, r, is 1 less than the term number that corresponds to it. Basedfrom the above, if the 6th term is to be found, then the 6th term, a6 is equal to the 1stterm, a1, multiplied by the common ratio r raised to the 5th power, or, in symbolsSo that the nth term is a6 = a1 ⋅ r 5 = a1 ⋅ r 6−1 an = a1 ⋅ r n−1 Rule or Formula for the General Term of a Geometric Progression: If an is a geometric sequence with common ratio, r, then an = a1 ⋅ r n−1 where n is the number of the term (term number) and a1 is the 1st term. A lot of problems involving geometric progression is solved using this rule for thegeneral term of a geometric progression. You will see in the following examples.Example 1. Find the 1st five terms of a geometric progression whose 1st term is 2 and whose common ratio is -3. Solution: Wait! This is the same problem as the one found in Example 2 of Lesson 1! Of course it is! But this time, it will be solved using the rule for the general term of a geometric progression. In lesson 1, each term after the 1st is found by finding the preceding term. Since a1 = 2 and r = -3, then proceed as follows an = a1 ⋅ r n−1 7

a2 = a1 ⋅ r 2−1 = a1 ⋅ r1 = 2(-3) = -6 a3 = a1 ⋅ r 3−1 = a1 ⋅ r 2 = 2(-3)2 = 2(9) = 18 a4 = a1 ⋅ r 4−1 = a1 ⋅ r 3 = 2(-3)3 = 2(-27) = -54 a2 = a1 ⋅ r 5−1 = a1 ⋅ r 4 = 2(-3)4 = 2(81) = 162The same first 5 terms of the sequence as in Example 2 of Lesson 1 are found. 2, -6, 18, -54, 162 An advantage of solving the problem using the rule for the general termof the geometric progression is that, you do not depend on the previous term toget the next term. The rule for the general term of a geometric progression is aconvenient way for you to find the nth term of a geometric progression.Example 2. Find the nth term of the geometric progression whose first three terms area. 4, 8 , 16 b. 5, -10, 20 39Solution: Since the general term of a geometric progression is an = a1 ⋅ r n−1 , you have to identify the 1st term and the common ratio. The 1st term is:a. a1 = 4 b. a1 = 5The common ratio is not given, so you have to find the common ratioby dividing a term by the preceding term. For this case, take a2 anda1 so that8a. r = a2 = 3 = 8 ÷ 4 = 8 ⋅ 1 = 2 b. r = a2 = − 10 = -2a1 4 3 34 3 a1 5Replace a1 and r into the rule for the general term:a. an = a1 ⋅ r n−1 b. an = 5 · (-2)n-1an = 4 ·  2  n−1 3Notice that it is not possible to simplify further 8

Try this outA. Write the 1st five terms of the geometric progression with the given 1st term and common ratio.1. a1 = 5 r=22. a1 = 3 r=43. a1 = 3 r = -0.54. a1 = -35. a1 = .5 r = -2 r = 0.5B. Find the nth term of the geometric progression.1. 2, 8 , 32,…2. -4, 12, -36,…3. 6, 4, 8 ,… 34. -6, 5, − 25 ,… 65. 9, -3, 1,…6. 1, 5, 25,…7. -3, 6, -12,…8. 8, 6, 9 ,… 29. -2, 4 , − 8 ,… 3910. 8, − 4 , 2 ,… 3911. 3 , 3 , 3 ,… 100 10000 100000012. 0.5, 0.05, 0.005,…13. 0.45, 0.0045, 0.000045,…14. 1, -x, x²,…15. c², c5, c8,… 9

Lesson 3 Applications of the General Term of a Geometric Progression You are now ready to apply the formula of a geometric progression. As you goover the examples, the specific skills will be identified to you.Finding the Specific Term of a Geometric ProgressionExample 1. Find the sixth term of the geometric progression 3, 6, 12,… Solution: Find the common ratio. r = a2 = 6 = 2 a1 3 Substitute in the formula for the nth term of a geometric progression with n = -6, a1 = 3, r = 2. an = a1 ⋅ r n−1 a6 = 3 · (2)6-1 = 3(2)5 = 3(32) = 96 Notice that the problem simply asks for only the 6th term. You can solve this bylisting all the terms of the sequence. So that continuing the sequence will give the 6thterm. a1, a2, a3, a4, a5, a6 ↓↓ ↓ ↓ ↓↓ 3, 6, 12, 24, 48, 96 So the 6th term is 96.Example 2. Find the 7th term of the geometric progression whose 1st term is 4 and whose common ratio is -3. Solution: Since a1 = 4, r = -3 and the 7th term is asked, then the term number, n = 7. So that, an = a1 ⋅ r n−1 a7 = a1 ⋅ r 7−1 a7 = 4(-3)6 10

a7 = 4(729) = 2916Finding a Term, Given Two Other Terms of a Geometric ProgressionExample 3. The third and sixth term of a geometric progression is 5 and -40 respectively, find the eighth term.Solution: Notice that neither the1st term nor the common ratio is given in this problem. So the solution for this one is not the conventional way of solving geometric progressions. Start from what is given. Since a3 = 5 and a6 = -40, then n = 3 and n = 6, respectively. From the formula for the nth term of a geometric progression,a3 = a1r3-1 an = a1 ⋅ r n−1 a6 = a1r6-1 (2)a3 = a1r2 and a6 = a1r55 = a1r2 -40 = a1r5 (1)Now a system of equations in two variables results. Recall that one wayof solving a system is by substitution. To do this, solve one variable,in terms of the other. In this case, solve for a1 in terms of r, in equation (1).So that, 5 = ar² 5 = a1 ⋅ r 2 Divide both sides by r2. r2 r2 5 = a1 r2Then, replace a1 by 5 in equation (2) r2 -40 = a1 ⋅ r 5 -40 =  5  ⋅ r 5 simplifying r 5 becomes r3.  r2  r2 − 40 = 5r 3 divide both sides by 5. 55 since -8 is the third power of -2. -8 = r³ r = -2 11

Substitute -2 for r in any of the two equations to solve for the othermissing variable, a1. Using equation (1),5 = a1r² (1)5 = a1 · (-2)² Divide both sides by 4.5 = 4 · a1 5 = 4a1 44a1 = 5 4Finally, solve the problem, that is, find the 8th term:an = a1 ⋅ r n−1a8 = a1 · r 8−1a8 =  5  ⋅ (− 2)7 4a8 =  5  ⋅ (− 128) = 5(-32) = -160 4 There is another way of solving the problem above without using systems ofequations. It is given below.Solution 2: Since the geometric progression gives the 3rd and 6th terms, it can be written as__, __, 5, __, __, -40, __, __a1, a2, a3, a4, a5, a6 , a7, a8Deleting the first two terms, another geometric sequence is found thatbegins as5, __, __, -40, __, __a1, a2 , a3, a4 , a5, a6Note that this second geometric progression has the same common ratioas the original geometric progression. For this second sequence, a1 = 5and a4 = -40. Substituting the formula an = a1 ⋅ r n−1 , then, 12

a4 = a1 · r4 - 1 Divide both sides by 5. a4 = a1 · r³ Take the third root of each side.-40 = 5 · (r³)− 40 = 5r 3 55 -8 = r³ -2 = rSince what is asked is to find the 8th term in the original sequence and ithas become the 6th term in the second sequence, solve for the 6th termin the second sequence,a6 = a1r6-1a6 = a1r5a6 = 5(-2)5a6 = 5(-32)a6 = -160Therefore, the 8th term in the original sequence is -160. Notice that youdo not have to compute for the 1st term of the original sequencesince what is asked is only to find the 8th term.Finding the Term Number (Position) of a Term in a Finite Geometric SequenceExample 4. In the geometric progression whose 1st term is -5 and whose common ratio is -2, which term is 10,240?Solution: It is given that a1 = -5, r = -2 and 10,240 as the nth term. Substituting in the formula for the general term of a geometric progression gives an = a1 ⋅ r n−1 Since an – 1 is the same as a n .10,240 = -5 · (-2)n-1 a1 10,240 = − 5 ⋅ (−2) n−1 −5 −5 -2048 = (-2)n-1 -2048 = (−2) n (−2)1 13

(−2)− 2048 = (−2) n (−2) Multiply each side by (-2).  −2  Since 4096 is the 12th root of (-2).4096 = (-2)n n = 12Therefore 10240 is the 12th term. Notice that the terms in a geometric progression get quite large early in thesequence where the common ratio is greater than 1. The problem above can then besolved by listing down all the terms until the needed term comes out.Solution 2: Since the 1st term is -5 and the common ratio is -2, then by listing the terms of the problem above,-5, 10, -20, 40, -80, 160, -320, 640, -1280, 2560, -5120, 10240a1 a7 a12Stop at 10,240 and count the number of terms. So 10,240, is the 12th term.Example 5. Find the common ratio of a geometric sequence if the first term is 1 and 2 the eighth term is 2187 . 2Solution: Since a1 = 1 , and a8 = 2187 then replacing them in the formula for 22 the general term of a geometric progression, an = a1 ⋅ r n−1 a8 = 1 ⋅ r 8−1 2 2187 = 1 ⋅ r 7 22  2187 = r7  Multiply both sides by 2. 2 2 2 2 Since 37 = 2187.   2187 = r7 7=rTherefore, the common ratio is 7. 14

Try this outA. Find the indicated term for each geometric sequence.1. 2, 10, 50,… a102. -1, -3, -9,… a153. 1 , 1 , 1 ,... a12 2 6 18 a184. 2 , − 1 , 1 ,.... a40 336 a95. a1 = 5, r = − 1 a8 5 a76. 2, 8, 32,…. a77. 4, 3, 9 ,.... a9 a8 48. 6, -4, 8 ,.... 39. -5, 15, -45,….10. 1, 2,2,....11. 3, 3 3,9,....B. Solve as directed. 12. Find the 8th term of the geometric progression 8, 4, 2, 1,… 13. Find the 6th term of the geometric progression whose first two terms are 4 and 6. 14. Find the 10th term of the geometric progression whose 5th term is 48 and 8th term is -384. 15. Find the 1st term in the geometric progression where the 4th term is 4 and the 7th term is 32. 16. In the geometric progression 4, 64, 1 024,… which term is 262 144? 15

Lesson 4 Geometric Means When the first and the last terms of a geometric sequence are given, the termsbetween them are called the geometric means. For example, the 3 geometric means ofthe geometric sequence 2, 6, 18, 54, 162 are 6, 18 and 54. To solve for the geometric means of a given geometric sequence, the formula forthe nth term of a geometric sequence is also used. Study the examples below.Example 1. Insert 3 geometric means between 4 and 324.Solution: Listing down the geometric sequence will show that there are five terms, which means that n = 5. So that a5 = 324 and a1 = 4. 4, __, __, __, 324Substituting in the formula:an = a1 ⋅ r n−1 divide both sides by 4a5 = 4 ⋅ r 5−1 since 81 is the fourth power of ±3324 = 4 ⋅ r 4324 = 4 ⋅ r 4 4481 = r4±3=rNote that the common ratio, r, takes two values, +3 and -3. So that thereare two sets of geometric means that can answer the question. To getthe desired geometric means simply multiply the common ratio to the firstterm first and so on.For r = 3, 4, 12, 36, 108, 324For r = -3, 4, -12, 36, -108, 324The geometric means are therefore ±12, 36, and ±108. 16

Example 2. Insert four geometric means between 3 and 96.Solution: Listing down the sequence gives3, __, __, __, __, 96So that n = 6, a6 = 96 and a1 = 3. Substituting in the formula,an = a1 ⋅ r n−1 divide both sides by 3a6 = a1 ⋅ r 6−1 since 32 is the fifth power of 296 = 3 ⋅ r 596 = 3 ⋅ r 5 3332 = r5 2=rFor this one, since r is only 2 then only one set of answer is possible.3, 6, 12, 24, 48, 96The geometric means are 6, 12, 24 and 48.Example 3. Find the geometric mean between 12 and 192.Solution: Here only a single term is asked. So that12, __, 192Using the formula for the nth term of a geometric sequence,an = a1 ⋅ r n−1 divide both sides by 12192 = 12 ⋅ r 3−1 16 is the second power of ±4192 = 12 ⋅ r 2192 = 12 ⋅ r 2 12 1216 = r2±4 = r 17

Multiplying the first term, 12, by the common ratio , ±4, the computed geometricmean is either 48 or -48. There is another way of solving for the geometric mean between two terms.Study how it’s done below. Given the same problem, as in Example 3. First let the second term be x , thenthe geometric sequence becomes 12, x, 192. Remember in Lesson 1 the common ratiois found by dividing the second term by the first term and is equal to the third termdivided by the second term. That is, r = a2 = a3 a1 a2So that substituting a1 by 12, a2 by x and a3 by 192, x = 192 12 xSolving the proportion by using cross multiplication, gives x = 192 using the square root property. 12 x x 2 = 12 ⋅192 x 2 = 2304 x = ± 48Presto! We got the same answer as the one above in one step only.Generally, Geometric Mean Between Two Numbers If b, c and d form a geometric sequence then c is the geometricmean between b and d. So that, c=d c 2 = bd c=± b⋅d bc Therefore, the geometric mean between two terms/ numbers is thesquare root of the product of the two terms/ numbers. 18

Example 4. Find the geometric mean between the two numbers. a. 8 and 72 b. -7 and -112Solution: Substituting in the formula for the geometric mean between two numbers c=± b⋅d a. c = ± 8 ⋅ 72 = ± 576 = ± 24 b. c = ± (−7)(−112) = ± 784 = ± 28Try this outA. Find the geometric mean between the given two numbers.1. 4 and 20 6. 3 and 252. 8 and 128 2 25 7. -2 and -2/3 3 8. 2and3 23. 2 and 9. 30 and 240 44. 6 and 2165. 4 and 144 10. 3 and 20 57B. Do as directed. 1. Insert two geometric means between 15 and 15 . 8 2. Insert four geometric means between 4 and -972. 3. Find the geometric mean between 5 and 500. 4. Insert four geometric means between 25 and 8 . 4 125 5. Insert three geometric means between 2 and 27 . 38 19

Let’s Summarize1. A geometric sequence is a sequence where each term after the first is found by multiplying it by a common ratio, r. an = r · an-12. The common ratio is found by dividing any term by the preceding term.r = a2 = a3 = a4 = a5 =….= ana1 a2 a3 a4 a n −13. The nth term of a geometric sequence is given by the formula an = a1 ⋅ r n−1where a1 is the first term, r is the common ratio and n is the termnumber.4. This formula is used in findinga. the general term of a given geometric progression;b. the specific term of a given geometric progression;c. a term given two other terms of a geometric sequence;d. the term number of a term in a given geometric sequence; ande. geometric means between two terms of a geometric sequence.5. The terms found between two terms of a geometric sequence are called geometric means.6. The geometric mean between two terms of a geometric sequence is given by the formula c=± b⋅dwhere b and d are the two given terms of the geometric sequence. 20

What have you learnedA. Which of the following is a geometric progression? Write GP, if it is or not, if not. 1. 3, 6, 9, 12,… 2. 9, 3, 1,… 3. 1, 22, 32,... 4. 3.33, 2.22, 1.11,… 5. 3, 9 6 , 162,…B. Do as directed. 6. Find the common ratio of the geometric sequence 3, 6, 12, 24, …. 7. What is the general term for the geometric sequence in no. 1? 8. Find the 8th term of the sequence 2, 6, 18, … 9. Find the first 6 terms of a geometric sequence with a1= 25, r = − 1 . 5 10. Insert two geometric means between 28 and 224. 21

Answer KeyHow much do you know 6. 3 7. 5 ⋅ (3) n−11. GP 8. 6402. GP 9. 81, 27, 9, 1, 13. Not GP 34. GP 10. 14, 28, and 565. Not GP 6. r = 2 7. NotTry this out 8. 1Lesson 1 r= 5A. 1. r = 2 2. Not 9. r = -4 3. r = -3 10. Not 4. r = - 1 2 6. 2, -6, 18, -54, 162 7. 2 1 1 1 1 5. r = 3 ,- , ,- ,B. 3 3 6 12 24 1. 2, 6, 18, 54, 162 8. 1, 0.5, 0.25, 0.125, 0.0625 2. 3, 6, 12, 24, 48 9. -1, -0.5, -0.25, -0.125, -0.06253. 10, 5, 5, 5 5 , 10. -2, 4, -8, 16, -32 248 4. -3, 6, -12, 24, -484. 1 1 5. 0.5, 0.25, 0.125, 0.0625, 0.03125 32, 8, 2, , 4 165. 3, -6, 12, -24, 48Lesson 2A.1. -5, -10, -20, -40, -802. 3, 12, 48, 192, 7683. 3, -1.5, 0.75, -0.375, 0.1875 22

B. 9. -2  − 2  n−1 31. (2)2n-1 10. 8  − 1 n−12. -4(-3)n-1 63. 6  2  n−1 11. 3  1  n−1 3 100 4. -6  − 5  n−1 12. 5(0.01)n 6 13. 45(0.01)n5.  − 1 n−3 14. (-x)n-1 3 15. c3n-16. (5)n-1 9. -3 645 10. 167. -3(2)n-1 11. 81 38. 8  3  n−1 12. 1 4 16 13. 243Lesson 3 81. 2(5)9 = 3 906 250 14. -1 5362. -314 15. 13. 1  1 11 2 2  3 16. 5th term4. - 1 23 3(2)165. - 1 5 386. 131 0727. 2187 40968. 128 243

Lesson 4 6. ± 6/5A. 7. ± 2 31. ± 202. ± 32 3 8. ± 63. ± 3 9. ± 60 24. ± 36 10. ± 2 215. ± 24 7B1. 15 , 15 4. 5 , 1, 2 , 4 2 5 25 242. -12, 36, -108, 304 5. 1, 3 , 9 243. 50 6. r = 2What have you learned 7. 3(2)n-1 8. 4 3741. Not 9. 25, -5, 1, - 1 , 1 , - 12. GP3. Not 5 125 1254. Not 10. 56 and 1125. GP 24

Module 6 Radical Expressions What this module is all about This module is all about radical equations and verbal problems involvingradicals. It is assumed that you are already good in solving linear and quadraticequations since knowledge of these will help you a lot in this module. What are you expected to learn This module is made for you to: 1. solve radical equations; and 2. solve problems involving radical equations. How much do you knowA. Solve the following radical equations: 1. x = 5 2. x + 4 = 11 3. x = - 17 3 4. 6 – 2 3x = 0 5. 4x − 5 = x + 9 6. x + 7 = x – 5 7. x 2 + 5 – x + 2 = 0 8. 3x + 1 = 1 - x + 4B. Read the following word problems and solve what is asked for: 9. The square root of the sum of a certain number and 8 is 13. find the number. 10. The square root of the sum of 2 consecutive odd integers is 1 less than 3 times the smaller integer. Find the integers.

What you will do Lesson 1 Radical EquationsThe following are examples of radical equations: 2x – 4 = 7 x +1 = 2x − 5 A radical equation is an equation that has variables in one or moreradicands. To solve radical equations, one must first get rid of the radicand in theequation. This can be done by squaring both sides of the equation using theproperty: The Squaring Property of Equality If a = b, then a² = b²To understand the property, study the examples below.Example 1. Solve: x = 6 Solution: Since the unknown variable is inside the square root symbol, then apply the squaring property. So that: x =6 ( x )2 = (6)2 x = 36Check: x =6 36 = 6 6=6 The solution checks.Example 2. Solve the equation: x + 2 = 3 Solution: Because the equation contains a radical, square both sides to eliminate it. Then proceed as follows: 2

x+2 =3 square both sides simplify ( x + 2 )² = 3² x+2=9 x=7Check: x+2 =3 7+2 =3 substitute x by 7 9 =3 3=3 The solution checks. Note that from the previous examples the squaring property is easily usedsince the radicals are isolated on one side of the equation. Now, what happens if both sides of the equation contain radicals as in thenext example?Example 3. Solve 3x + 2 = x + 10Solution: In this example the radicals on both sides of the equation may be eliminated by using the squaring property right away! ( 3x + 2 )2 = ( x + 10 )² square both sides simplify 3x + 2 = x + 10 2x = 10 –2 2x = 8 x= 8 2 x=4Check : 3x + 2 = x + 10 3(4) + 2 = 4 + 10 substitute x by 4 12 + 2 = 14 14 = 14 The solution checks. From the three examples, you can say that it’s easy to solve radicalequations. Notice how the equations are simplified by squaring both sides whenthe radicals containing the unknown is isolated on one side of the equation as inExamples 1 and 2. Also, when the radicals are on both sides of the equation asin Example 3. 3

Example 4. Solve the equation: x + 1 - 5 = 3 Solution: Notice that the left-hand side consists of two terms, the first being a radical, x + 1 and the second, a constant, -5. We cannot square both sides applied right away. You must rearrange the terms first to isolate the radical on one side of the equation. So that, x +1 - 5 = 3 square both sides x +1 = 3 + 5 x +1 = 8 ( x + 1 )2 = 82 x + 1 = 64 x = 64 – 1 x = 63Check: x +1 - 5 = 3 63 + 1 - 5 ? 3 64 - 5 ? 3 8–5?3 3=3 The solution checks.Example 5. Solve the equation: x +1 + 5 = 3Solution: Rearrange the terms first to isolate the radical on one side of the equation. So that, x +1 + 5 = 3 x +1 = 3 - 5 x + 1 = -2 You may already stop here. Notice that the left- hand side, x + 1 willnever be negative. Whereas, the right- hand side is negative. This makes theequation x + 1 = -2 not trueBut, just then, suppose this step is missed? 4

Let’s continue the solution: ( x + 1 )² = (-2)² square both sides x+1=4 simplify x=3Check: x +1 + 5 = 3 3+1 + 5 = 3 4 +5=3 4 +5=3 2+5=3 7≠3 Because 7 = 3 is a false result, 3 is not a solution! So that the originalequation has no solution. Truly the equation, as you’ve seen above, has nosolution. Example 5 illustrates that squaring both sides of an equation can lead tosolutions, called extraneous solutions. These kind of solutions do not satisfythe original equation. This is because the squaring property may not produceequivalent equations. When both sides of an equation are squared, the newequation may have solutions that the first one does not have.For example, the equation x=1 (1)has just one solution, 1. Squaring both sides, you will get x² = 1 (2)which has two solutions, 1 and –1. Thus, the equation x = 1 and x² = 1 do nothave the same solutions and so are not equivalent. Whereas, it is true that anysolution of equation (1) is a solution of equation (2), it is not true that any solutionof equation (2) is a solution of equation (1). So that, it is important that you checkthe solutions. There are times when a quadratic equation results in using the squaringproperty in a certain radical equation as will be seen in the next example.Example 6. Solve: x–5= x+7Solution: x- 5 = x + 7 5

(x – 5)² = ( x + 7 )² use the squaring property expand or simplify x² - 10x + 25 = x + 7 factor x² - 11x + 18 = 0 use the zero product property (x – 9) (x - 2) = 0 x - 9 = 0 or x - 2 = 0 x=9 x=2Check: if x = 9 x–5= x+7 if x = 2 9–5= 9+7 4 = 16 4=4 x-5= x+7 2-5= 2+7 -3 = 9 -3 ≠ 3 9 is the solution, not 2. The next examples involve finding the square of a binomial on the otherside of the equation.Example 7. Solve: 3x + 4 = 8 – xSolution: 3x + 4 = 8 – x ( 3x + 4 )2 = (8 – x)2 3x + 4 = 64 – 16x + x2 square both sides the right-hand side is the result of 0 = 64 – 4 - 3x – 16x + x2 squaring the binomial (8 – x)2 0 = 60 – 19x + x2 combine similar terms 0 = (x – 15)(x – 4) x – 15 =0 or x – 4 = 0 factor x = 15 or x = 4Check: if x = 15 3x + 4 = 8 – x 3(15) + 4 ? 8 – 15 If x = 4 49 ? -7 7 ≠ -7 3x + 4 = 8 – x 3(4) + 4 ? 8 – 4 16 ? 4 4=4 6

The solution is 4.Example 8. Solve: 3 + 27 − 3x = xSolution: The squaring property cannot be used right away since the radical is not yet isolated, so that 3 + 27 − 3x = x 27 − 3x = x – 3 subtract 3 to isolate the radical square both sides ( 27 − 3x )² = (x – 3)² expand or simplify combine similar terms 27 – 3x = x² - 6x + 9 factor 0 = x² - 6x + 9 + 3x – 27 use the zero product property 0 = x² - 3x – 18 0 = (x – 6)(x + 3) x – 6 = 0 or x + 3 = 0 x=6 x = -3Check: if x = 6 3 + 27 − 3x = x 3 + 27 − 3(6) ? 6 3 + 27 − 18 ? 6 3+ 9?6 3+3?6 6=6 if x = -3 3 + 27 − 3x = x 3 + 27 − 3(−3) ? -3 3 + 27 + 9 ? -3 3 + 36 ? -3 3 + 6 ? -3 9 ≠ -3Clearly, 6 and –3 is an extraneous solution that should be discarded. Suppose in Example 8, the squaring property is applied without isolatingthe radical: (3 + 27 − 3x )² = x² 9 + 2(3) 27 − 3x + ( 27 − 3x )² = x² 9 + 6 27 − 3x + (27 - 3x) = x² 6 27 − 3x = -9 – 27 + 3x + x2 6 27 − 3x = x2 + 3x – 36 7

Presto! What has resulted is a more complicated expression than the originalequation.Remark: When a radical equation requires squaring a binomial on one side of the equation, be careful not to square only the binomial in the product but also the middle term which is twice the product of the first and last terms of the binomial. Now, there are times when the squaring property has to be applied morethan once, as you will see in the next example.Example 9. Solve: x - 1 = x − 5Solution: In this one, there is no way to isolate the radical since there is a radical on both sides of the equation. Thus, x -1= x−5 square both sides ( x - 1)² = ( x − 5 )² expand and simplify ( x )² - 2 x + 1 = x – 5 simplify x–2 x +1=x–5 isolate the radical -2 x = -6 square both sides x =3 ( x )² = 3² x=9Check: x -1= x−5 9-1? 9−5 3 -1 ? 4 2=2 The solution 9 checks.The examples above all follow the steps identified below:Steps in solving radical equations:1. Whenever possible, rearrange the terms to isolate a single radical on one side of the equation2. Square both sides of the equation.3. If a radical term remains, do steps 1 and 2 again. Solve for the unknown.4. Check the solution in the original equation. This step is required. 8

Try this outA. Check to see if the given value for x is a solution of the equation.1. 3x + 18 = x a. 6 b. -32. 2x + 3 = 0 a. 3 b. -13. 3 3x + 4 = 3 2x − 1 a. -3/5 b. -54. Is 16 a solution of the equation x = -4? If not, what is the solution of this equation?5. What is wrong with the first step in the process? 2x − 3 = x – 2 2x – 3 = x2 + 4B. Solve each equation. Check all solutions.1. x = 7 21. 2x + 10 + 3 = 52. x = -9 22. 9x + 25 - 2 = 33. x = -1 23. 7 − 5x + 4 = 34. x = 3/2 24. 7 + 6x - 4 = -35. x + 3 = 2 25. 3x + 3 = 3 x − 16. x − 2 = 3 26. 2 3x + 4 = 5x + 97. x − 5 = 5 27. 3x + 6 = 2 2x − 118. x + 8 = 12 28. 10 − 3x = 2x + 209. 3 − x = -2 29. 2 4x + 5 = 5 x + 410. 5 − x = 10 30. 2 9x + 16 = 3x + 6411. 6 + 2x = 4 31. x + 1 = x - 112. 7 + 3x = -4 32. x − 1 = x + 113. 5x − 5 = 5 33. x - 1 = x − 114. 6x + 19 = 7 34. x + 4 = x - 215. 4x − 3 = 3 35. x + 9 = x + 716. 11x − 2 = 3 36. x – 2 = x + 1017. 13x + 14 = 1 37. 7x + 2 - 2x = 018. 8x + 9 = 1 38. x = 3 − x + 319. x + 3 + 5 = 12 39. 3x + 3 + 5 = x20. 5x + 9 + 4 = 7 40. x = x − 4 + 4 9