Math G9 Quarter 2 Variation PDF

QUARTER 2 Mathematics G9  Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work f...

QUARTER 2 Mathematics G9  Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this book are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and the authors do not represent nor claim ownership over them. This module was carefully examined and revised in accordance with the standards prescribed by the DepEd Regional Office 4A and CLMD CALABARZON. All parts and sections of the module are assured not to have violated any rules stated in the Intellectual Property Rights for learning standards. The Editors PIVOT 4A CALABARZON Math G9 PIVOT 4A Learner’s Material Quarter 2 Second Edition, 2021 Mathematics Grade 9 Job S. Zape, Jr. PIVOT 4A Instructional Design & Development Lead Jisela N. Ulpina Content Creator & Writer Jhonathan S. Cadavido Internal Reviewer & Editor Lhovie A. Cauilan & Jael Faith T. Ledesma Layout Artist & Illustrator Jhucel A. del Rosario & Melanie Mae N. Moreno Graphic Artist & Cover Designer Ephraim L. Gibas IT & Logistics Published by: Department of Education Region IV-A CALABARZON Regional Director: Francis Cesar B. Bringas PIVOT 4A CALABARZON Math G9  Guide in Using PIVOT 4A Learner’s Material For the Parents/Guardians This module aims to assist you, dear parents, guardians, or siblings of the learners, to understand how materials and activities are used in the new normal. It is designed to provide information, activities, and new learning that learners need to work on. Activities presented in this module are based on the Most Essential Learning Competencies (MELCs) in Mathematics as prescribed by the Department of Education. Further, this learning resource hopes to engage the learners in guided and independent learning activities at their own pace. Furthermore, this also aims to help learners acquire the essential 21st century skills while taking into consideration their needs and circumstances. You are expected to assist the children in the tasks and ensure the learner’s mastery of the subject matter. Be reminded that learners have to answer all the activities in their own answer sheet. For the Learners The module is designed to suit your needs and interests using the IDEA instructional process. This will help you attain the prescribed grade-level knowledge, skills, attitude, and values at your own pace outside the normal classroom setting. The module is composed of different types of activities that are arranged according to graduated levels of difficulty—from simple to complex. You are expected to : a. answer all activities on separate sheets of paper; b. accomplish the PIVOT Assessment Card for Learners on page 38 by providing the appropriate symbols that correspond to your personal assessment of your performance; and c. submit the outputs to your respective teachers on the time and date agreed upon. PIVOT 4A CALABARZON Math G9  Parts of PIVOT 4A Learner’s Material K to 12 Learning Descriptions Delivery Process This part presents the MELC/s and the desired Introduction What I need to know learning outcomes for the day or week, purpose of the lesson, core content and relevant samples. This maximizes awareness of his/her own What is new knowledge as regards content and skills required for the lesson. This part presents activities, tasks and contents What I know of value and interest to learner. This exposes Development him/her on what he/she knew, what he/she does What is in not know and what he/she wants to know and learn. Most of the activities and tasks simply and directly revolve around the concepts of What is it developing mastery of the target skills or MELC/s. In this part, the learner engages in various tasks What is more and opportunities in building his/her knowledge, skills and attitude/values (KSAVs) to meaningfully connect his/her concepts after Engagement doing the tasks in the D part. This also exposes What I can do him/her to real life situations/tasks that shall: ignite his/ her interests to meet the expectation; make his/her performance satisfactory; and/or produce a product or performance which will help What else I can do him/her fully understand the target skills and concepts. This part brings the learner to a process where he/she shall demonstrate ideas, interpretation, What I have learned mindset or values and create pieces of Assimilation information that will form part of his/her knowledge in reflecting, relating or using them effectively in any situation or context. Also, this What I can achieve part encourages him/her in creating conceptual structures giving him/her the avenue to integrate new and old learnings. This module is a guide and a resource of information in understanding the Most Essential Learning Competencies (MELCs). Understanding the target contents and skills can be further enriched thru the K to 12 Learning Materials and other supplementary materials such as Worktexts and Textbooks provided by schools and/or Schools Division Offices, and thru other learning delivery modalities, including radio-based instruction (RBI) and TV-based instruction (TVI). PIVOT 4A CALABARZON Math G9 WEEKS Variation 1-2 Lesson I There are varied things around you that you should know how things are related to each other. Like how is time related to the speed of a vehicle or even as simple as the relationship of the length of your arm span to your height. Quantities may vary directly, that is as one quantity increases the other quantity must also increase or they may vary inversely, that is if one quantity increases the other quantity decreases. When two quantities increase at the same time or decrease at the same time, it shows direct variation. However, if one quantity increases and the other decreases, it shows inverse variation. You encounter such situations in your everyday life. For instance, walking slowly going to school takes longer time, but walking fast takes shorter time reaching the school. Your speed in walking varies inversely with the time. The faster you walk the shorter the time it takes you to reach your destination. Another situation is when your mother asked you to buy rice, which cost ₱55.00 a kilo. If one kilo costs ₱55.00, how much will 10 kilos of rice cost? In this situation the more kilos of rice you buy, the more money you will pay. Hence, this is an example of direct variation. Situations involving 3 or more quantities that varies from one another may show joint variation or combined variation. D Direct Variation In your previous lesson you were able to represent relationship between two quantities using graphs, table of values and equations. Examples: 1. Given: Graph: Table of values: x 1 2 3 4 5 y 2 4 6 8 10 Equation: y = 2x You can see from the table of values that as the value of x increases the value of y also increases. The graph is a line that rises to the right. This is an illustration of a direct variation. This means that y varies directly as x. In symbols: y y  k or y = kx or x  , where k is the constant of variation or x k constant of proportionality. In this case the constant of variation is 2. PIVOT 4A CALABARZON Math G9 6 2. You will be celebrating your 14th birthday and you want to have 14 balloons in your birthday party. The table below shows the number of balloons (x) and the corresponding cost (y). As you can see, as the values of x increase, the x 2 4 6 8... 14 values of y also increase. Hence, the cost of balloons y 10 20 30 40... 70 varies directly to the number of balloons. y1 y2 10 20 Since y varies directly as x, then the constant of variation k: x  x  2  4  5 1 2 The constant of variation (k) is 5. Thus, the equation of the direct variation is y = 5x. 3. y varies directly as x. If y = 12 when x = 4, A. Find the constant of variation C. What is x when y is 36? B. What is the equation? Solution: y 12 A. k   3 C. using the equation of variation: y = 3x x 4 36 B. y = 3x 36 = 3x x=  12 3 When the value of y increases from 12 to 36, the value of x increases also from 4 to 12. Direct Square Variation The relationship between the area (A) and the square of the radius (r 2) of a circle is an example of a direct square variation. A = πr2, the constant of variation is π or 3.1416. Direct square variation states that if y varies directly as the square of x, there is y y y a nonzero constant k such that  k or y = kx2 or x2  where, x. x2 k k The graph of direct square variation is quadratic in nature. It follows a quadratic curve. Examples: 1. y varies directly as the square of x. If y = 27 when x = 3, find x when y = 81. Solution: Find the constant of variation: y 27 27 k   2  3 x2 3 9 Equation of variation: y = 3x2 Find x when y = 81. Using the equation y = 3x2 , find x 81 81 = 3x2 x2 =  27 x= 27  3 3 3 2. If p varies directly as the square of q and p = 256 when q = 8, find: A. k B. equation C. p when q = 2 3 p 256 Solution: A. k  q 2  82  4 B. p = 4q2 C. p = 4q2 p = 4( 2 3)2 = 4(12)= 48 7 PIVOT 4A CALABARZON Math G9 Inverse Variation In real life, there are different situations that shows inverse relation. Inverse variation states that y varies inversely as x or y is inversely proportional k k to x if there is nonzero constant k, such that y  , k = xy or x ... The x y decrease of one quantity results to the increase of the other quantity. Examples: 1. The table below shows the number of persons (x) working together and the number of hours (y) the job is completed. Find the constant of variation and equation. Graph the given table of values. 2 2 x 1 2 3 4 5 Constant of variation: xy = 5   3   2(1)  2 5 3 1 2 2 y 2 1 2 Equation of variation: xy = 2 or y 3 2 5 x Graph: 2. If y varies inversely as x and y = 20 when x = 8, find A. constant of variation B. variation equation C. y when x = 16 Solution: A. k = xy k = 8(20) = 160 160 B. xy = 160 or y = 160 160 x C. y =   10 x 16 As the value of x increases from 8 to 16, the value of y decreases from 20 to 10. 3. The rate (r ) of the car varies inversely with the time (t). The car travels at a rate of 60 kph for 1.5 hours. Find the rate of the car when it took 1 hour to travel the same distance (d). Solution: In this case, the faster the car travels, the shorter time it will take to cover the same distance. Hence, the constant of variation here is the distance. Therefore, d = rt = 60(1.5) = 90 km. To solve for the rate : r  d  90  90 kph. t 1 The faster the car travels, the shorter the time it takes to travel the same distance. PIVOT 4A CALABARZON Math G9 8 Learning Task 1: A. Complete the table if y varies B. Complete the table if y varies directly as x or x2. inversely as x. y x k Equation y x k Equation 30 5 5 6 8 10 4 36 50 18 y = 3x 20 y= x 24 72 y = 2x2 12 y= x Joint and Combined Variation Joint and combined variation involves 3 or more quantities that may vary directly and or inversely to each other. Joint Variation If the ratio of one quantity to the product of the other two quantities is con- stant, the they vary jointly. That is, if y varies jointly as x and z, then y = kxz , y y y or  k. Or x  or z  xz kz kx Examples: 1. The surface area of a cylinder varies jointly as the radius and height. A = krh, where A is the surface area , r is the radius and h is the height of the cylinder. The constant of variation is k = 2π. Hence, A = 2πrh. Any of the variable r or h increases the area also increases. 2. Translate the following into variation equation: (a) p varies jointly as q and r. Equation: p = kqr (b) area of parallelogram varies jointly as its base and altitude. Equation: A = bh, k = 1 3. If m varies jointly as p and q and m = 50 when p = 5 and q = 2, find m when p = 10 and q = 6. Solution: Determine the constant of variation: y 50  k k 5 xz 5(2) m = kpq = 5(10)(6) = 300. As p and q increase m also increases. Combined Variation This is a variation where one quantity varies directly to other quantity and inversely to the other quantity. vx vx kw k  kw x w The equation v  x , w , v , or k means that v varies directly as w and inversely as z. Examples: 1. v varies inversely as x and directly as w. If v = 12 when x = 4 and w = 8, find w 9 PIVOT 4A CALABARZON Math G9 Solution: kw k (8) 12( 4) Equation: v =. Find k: 12  k= 6 x 4 8 6w 6w 24(3) Solve for w: v = 24  w  12 x 3 6 2. If p varies directly as the square of q and inversely as the square root of r, and p = 20 when q = 2 and r = 64, find p when q = 8 and r = 144. Solution: kq 2 k (2) 2 20 64 20(8) Equation: p = Solve for k: 20  k= 22  4  40 r 64 2 Solve for p: p = 40(8)  40(64)  640 144 12 3 Learning Task 2 A. Supply the missing value if y varies B. Supply the missing value if y varies jointly as x and z. directly as v and w and inversely as z. y x z k Equation y v w z k Equation 30 3 5 5 12 14 24 80 4 Y= 2xz 18 4 8 10 3vw 6 8 4 16 24 6 y  2z 60 10 Y =3xz 20 6 2 8vw y 3 z 48 12 Y= xy 4 12 7 15 6 E Learning Task 3 A. Identify if the given equation is a direct, inverse, joint or combined variation with k as the constant of variation. kn 1. b = kd 3. m = p 5. mn = kpq a 2. y = klm 4.  kc b B. Write the equation of variation and solve for the constant of variation. 1. p 1 2 3 4 5 3. p 3 6 12 24 48 q 7 28 63 112 175 q 16 8 4 2 1 2. p 3 6 12 24 48 4. p 6 12 18 24 30 q 16 8 4 2 1 q 2 4 6 8 10 PIVOT 4A CALABARZON Math G9 10  A A. Solve the following: 1. The distance (d) from the center of the seesaw varies inversely as the weight (w) of a person. JB who weighs 50 kg sits 3 feet from the fulcrum. How far from the fulcrum must JP sit in order to balance with JB if he weighs 35 kg? 2. The number of pages (p) that Ethan reads varies directly as the number of hours (t) he is reading. A. write the variation equation B. If he can read 21 pages in 14 minutes, how may pages can he read in 21 minutes? 3. The pressure of the gas is directly proportional to the temperature and inversely proportional to its volume. A. Write the variation equation. B. What happened to the pressure if the volume is reduced to half and the temperature is doubled? pq 2 4. Given the equation y  k , where k is the constant of variation, r determine which statement is true or false. A. y and r varies directly. B. y and q2 are directly proportional C. y and pq2 varies jointly D. p and r are inversely proportional E. y and p varies directly. 5. The volume of a cylinder is given by the formula V = πr2 h. If r is increased by 50% and the height is reduced by 25%, what will happen to the volume? What is the constant of variation? B. Give at least three examples of quantities or situations you know or you have experienced that show different types of variation. 11 PIVOT 4A CALABARZON Math G9 WEEK Integral and Zero Exponents 3 I Lesson You have learned in previous lessons about laws of exponents. Let us recall these laws: (a) Product of a Power: am · an = am+n am (b) Quotient of a Power:  amn an (c ) Power of a Power: (am)n = amn (d) Power of a product: (a · b)m = am bm m am (e )Power of a Quotient:    m a b b In the process of performing the laws of exponents, it may result to a positive, negative, or zero exponents. In this lesson, you will learn how to simplify expressions with integral (positive or negative integers) or zero exponent applying the different laws of exponents. D Zero Exponent a2 aa Simplify the expression. Write this expression in factored form aa 1 a2 a2 Applying the laws of exponent, quotient of a power,  a 2  2  a 0  1. Thus, for a2 any variable/number a , where a ≠ 0, a0 = 1. Examples: 1. 12x0 = 12(1) = 12 Only variable x is raised to zero power. 2. (12x)0 = (12)0 (x)0 = 1(1) = 1 Applying the law of power of a product, both 12 and x are raised to zero power. 3. (x0 + 3)(x + 3)0 = (1 + 3)( 1) = 4 The expression (x + 3) is raised to zero power it is equal to 1. 4 2 45a b 4. Simplify: Apply laws of exponent in simplifying. 5a 4 b Solution: 9a4 - 4 b2 - 1 = 9a0 b = 9b 5. Simplify: 2a0 + (2a)0 + 20a Solution: 2(1) + 1 + 1(a) = 2 + 1 + a = 3 + a 6. Simplify: 4m(n  p ) 0 4m 0 4m(1) Solution: m 4(1) PIVOT 4A CALABARZON Math G9 12 Integral Exponent An expression has an integral exponent if the exponent is either positive or negative integers. The expressions like 3x, 5x-2 + 2, m-8 , are examples or expressions with integral exponents. am One of the laws of exponent is quotient of a power that is :  a mn an Apply this law of exponent in simplifying the following: 1. x4 You can use factoring to check the answer: xxxx  x 4 2  x 2  x  x  x2 x2 xx x2 2.  x 24  x 2 The exponent is negative. Use factoring to simplify the x4 2 x x expression: x 4  1  2. Hence the expression x -2 is x x x x x x 1 equal to x2. 1 The negative sign in the exponent means the reciprocal. Thus, a n  a n. Similarly if the expression with negative exponent is in the denominator, get its reciprocal 1 to express the exponent as positive. Therefore , a  n  a. n 3x3 y 6 3. Simplify: x5 y 4 Solution: Observe the exponents of each variable. Let the variable with greater exponent remain in its position but subtract the smaller exponent 3x3 y 6 3 y64 3y2 from the larger exponent: 5 4  53  x y x x2 4. Express all exponent to positive integers. (a) 28a 5 b 6 c 4a 0 b 2 4b 2   7a 5b 4 c 3 c2 c2 (b) (5 - 2)5 · 3-3 + (5 + 2)0 = (3)5 · 3-3 + 1 = 32 + 1 = 9 + 1 = 10 (c ) [(8)2]-1 ⋅ (10 - 8)14 ⋅ 4-2 (8)-2 · (2)14 · 4-2 ,apply the law of exponent power of a power (23 )-2 · 214 · (22 )-2 ,change to the same base so can apply the law of 2-6 · 214 · 2-4 exponent 2-6+14+(-4) = 24 = 16 ,apply the law of exponent product of a power (d ) (5 - 2)5 · 3-8 + (5 + 2)0 (3)5 · 3-8 + 1 3-3 + 1 ,apply product of a power 1 +1 ,get the reciprocal of the number with the 33 negative exponent to make the exponent positive 1 27 28   ,change to similar fractions before adding 27 27 27 13 PIVOT 4A CALABARZON Math G9  E Learning Task 1: Express the following expression into nonzero and nonnegative exponents. Simplify your answer. 1 24 x 8 y 4  (ab)  1. 7-1 6. 11.   6 x5 y 4  9   5x0  1 9 2. (14abc)0 7.   12.  y  x 2 120a 5b 6 c 5 12b 3 3. 10-9 8. 13. 12a 2b 0 c 2 a 4 (4 x) 0 y 5 z 2 1 4. 5(xy)0 9. 14. (234 xyz ) 0 a 5 n 2 5  (5 xy ) o  1 5. 015 10.   15.    10  2 A Find the single value of the following without exponent. 1 1. 2-2 · 3-1 11. ( 4) 3 7 2 2. 100 + 10 12. 41 9 3. 4(60) + 3(3-1 ) 13. 1 2 82 4. 5-1 - 2-3 14. 40 2 5. (120)- 10 15.  3  5 6. 24 + 2– 2 7. (5 + 5-2)0 (60 + 6)-2 8. 20 33 9. 5(-6)0 10. 10-3 ÷ 103 PIVOT 4A CALABARZON Math G9 14  WEEK Rational Exponent and Radical Expressions 4 I Lesson Your previous lesson is all about integral and zero exponents. This time, your lesson is on rational exponents and radicals numbers. When simplifying expressions with rational exponents you should remember the laws of integral exponents, the rule is the same. However, you need to remember the rules in adding fractions. In adding fraction, you can only add fractions if they are similar otherwise you change dissimilar fraction to similar fraction. Examples: Simplify the following expressions: 1 3 1 3  1. a 2 a 2  a 2 2 ,apply the product of a power law of exponent by adding the exponents of the same base 4 = a 2  a2 ,apply the rule in adding fraction then, simplify. 6 x   1 6  6 x  3 6 2. 3 ,apply the power of a power law of exponents by multiplying the exponents. Use the rule in multiplying fractions = (6 x) 2  6 2 x 2 ,apply power of a product law of exponent = 36x2 ,simplify. 1 1  4 2  4  2 3.   1 ,apply the quotient of a power law of exponents 9 9  2 2  1 2 2 = ,express 4 and 9 in exponential form 3  1 2 2 2 22 2 = 2  ,multiply the exponents, then simplify. 3 3 2 2 1 2 1 4. 12 3  12 2  12  3  2 ,apply the product of a power law of exponent.. 4 3 7 = 126  6  12 6 ,change the fractions to similar fraction then add. 6 1 12  1 1 = 6  12 6  12 6  12 6  1212 6 ,simplify by applying laws of exponents. Exponential expressions with rational exponents can be expressed into radical expressions. 15 PIVOT 4A CALABARZON Math G9  D Radicals An exponential number whose exponent is a rational number can be ex- pressed as radicals Examples: Exponential Radicals 1 Exponent radical sign 3 Base 8 3 index 8 radicand The denominator of the rational exponent becomes the index of the radical number. 3 8 This is read as “cube root of 8” Exponential Radical 122 3 2 12 3   12  3 This is read as the “square root of 12 raised to the 3rd power” When changing the exponential number with rational exponent whose denomi- nator is 2 to radical number you may or may not write the index 2. m The exponential number a , when changed to radical is n n am read as the nth root of a raised to m. The nth root of a number is a number that is taken n times as a factor of the radicand a. You can express exponential expression to radical expression and vice versa. Examples: 1 5 4 1. Change to radical expression. 1 5 = 4 5 (read as the 4th root of 5) 4 3 2. Write 4 2 in radical expression 42 3   4 3 ( read as the square root of 4 cube or 4 raised to the 3rd power) 4. Write  8  in exponential form 3 4  8   8 The index (3) is the denominator of the fractional exponent 4 4 3 3 and the power (4) is the numerator of the fractional exponent. PIVOT 4A CALABARZON Math G9 16 5. Change the expression 5 to exponential form. 32  15  5 32     32  ,the 5 root of 32 is equal to 32 raised to one fifth. th   Laws of Radicals Laws of radicals are similar to the laws of integral exponents. Let a, b, m, n are integers: n    n  1 n 1. n a n  a    an  a The nth root of a raised to the power of n is equal to a.    a   b   a 1 1 1 2. n n n  b n  ab  n  n ab The n th root of a times the nth root of b is equal to the nth root of the product of a and b. 1 1 n a a  n  a n a 3.  1    n The nth root of a divided by the nth root of b is b  n b b n b equal to the nth root of the quotient of a and b. 1 4. m 1  1 m 1 n m a  a n   a n   a mn  mn a The nth root of the mth root of a number a   is the rth root of a where, r = mn. Examples Apply the laws of radicals. 3  12    1 3 1. 3 3  12 3   12 3  12     2. 5 4 5 8  5 ( 4)(8)  5 32  2 Multiply if the indices are the same, just multiply the radicand, copy the common index, then simplify. The fifth root of 32 is 2, meaning, 2 is one of the five equal factors of 32. In this case, 25 = 32. Hence, 5 32  2 18 18 3.   9 3 Divide the radicands if the indices are the same, 2 2 then simplify by extracting the square root of 9. The 9  3 , since 32 = 9. 1 1 1  4. 3 64  (64) 2 3  (64)  6 64  2 6 Change the given to fractional exponent. Apply law of radical number 4, then simplify. 2 1 2 1 43 7  5. 3 12 2. 12  12  12  (12) 3 2 3 2  (12) 6  (12) 6  6 12 7  6 12 6  12  6 12 6  6 12  126 12 ,apply the laws of rational exponents and radicals to simplify the expression. Learning Task 1 Use the Laws of rational exponent to simplify: a b  1 3 1 1. 3.  4. 124 5. n  3 4 3 1 3 4 12 2 a  2 1     4 4 2. 1 3 3b 2  b 2 17 PIVOT 4A CALABARZON Math G9  E Learning Task 2 A. Change exponential expression to radical expression. 2 3  15 52  1. 1253 3.  4 1 5.  5 5  16  2         1 2 2. 5x 2 4. xy 5 B. Write each expression in exponential form. 3 1. ( 7 ) 5 3.  5 4m  8 5. 83 4 11x  4 2a  3 4 228 A A. Write the laws of Radical Exponents and give examples. B. Apply the laws of rational exponents in simplifying the expression. 1 6.  16m n 3 9 3 1.   2 6    1  43 3  3 x  3 2. p q  7. 16   y8 8   2  12 14  2a  2a  1 1 3.  5 x   8. 2 2 2 3   1  a12  6 x    1 1 4.  6  9. 10 5  x4 2 b  1 2  5 1   4m 8 y 12 4  3q 2 r 4  5.  20   10.  1   8p   62    PIVOT 4A CALABARZON Math G9 18  WEEKS Simplifying Radical Expressions 5-6 Lesson I A radical is in its simplest form if: 1. The radicand does not contain a perfect square, cube or nth power. Examples: 2 is in its simplest form since 2 is not a perfect square, meaning the square root of 2 is not exact. 3 5 is in its simplest form since 5 is not a perfect cube, meaning, the cube root of is not exact. 2. There is no fraction in the radicand or no radical in the denominator. Example:  3  3 3 , the fractions is simplified such that     4  4 2   there is no radicand in the denominator. 3. The index of the radical is the lowest possible index. 6 Example: 8 is not yet in simplest form since the index 6 can be expressed as the product of 3 and 2. To simplify, express it as 3 8  2 since the cube root of 8 is 2 1 1 1 1    1 and square root of 2 is not exact or 3 8  (8) 2 3  (8) 6  23 6  2 2  2. You have to simplify radical expressions with: 1. a perfect nth factor; 2. a fraction radicand or denominator; and 3. reducible index. Recall that : a. n am is a radical expression which indicates the m nth root of am ; a  n am n b. n is the index which gives the order of radicals; and c. am is the radicand, the number within radical sign. D The properties of radicals which can be useful in simplifying radical expressions are as follows: 1. n ab  n a  n b The nth root of a product is equal to the nth root of each factor. 2. n m p  nm p The nth root of the mth root of a number is equal to the nmth root of the number. p n p  3. n q n q The nth root of the quotient is equal to the nth root of the numerator and the nth root of the denominator. 19 PIVOT 4A CALABARZON Math G9 Simplifying Radicals by Reducing the Radicand Reducing radicand is finding a factor of a radicand whose indicated roots can be found. Examples: 1. Simplify 50. Solution: 50 is not a perfect square, therefore you have to find a factor of 50 which is a perfect square. 50  25 .2 The factors of 50 are 25 and 2. 25 is a factor which is a perfect square, meaning you can extract the square root.  25  2 The square root of the product is equal to the square root of the factors. 5 2 The square root of 25 is 5 , square root of 2 is already in simplified form. Therefore, 50  5 2 is in simplest form. 2. Simplify 3 81. Solution: The factor of 81 which is a perfect cube is 27. 3 81  3 27  3  3 27  3 3  33 3 3 81  33 3 since, 33 3 3  33  33   27(3)  81. 1 3 Therefore,     7 5 3. Simplify 5 32 x y z. Solution: To get the root of variables with integral exponent, the exponent must be divisible by the index. 5 32 x 7 y 5 z  5 32 x 5 y 5  x 2 z  5 32 x 5 y 5  5 x 2 z  2 xy 5 x 2 z Take note that when you divide the exponent of x which is 7 by the index 5 there is a remainder of 2 and the exponent of z which is 1 cannot be divided by 5. Hence, the variables that will remain under the radical sign are x 2 and z. 4. Simplify 54 48a 4 b 5 c 6. Solution: Factor the radicand such that one factor is the 4th power of a certain number and the variable such that, the exponent is divisible by the index 4. 54 48a 4b10 c 5  54 16a 4b 8 c 4  3b 2 c  54 16a 4b 8 c 4  4 3b 2 c  5(2ab 2 c)4 3b 2 c  10ab 2 c 4 3b 2 c The factors of 48 are 16 and 3 where 16 is a perfect 4th power of a num- ber. The factors of b10 are b8 where the exponent is divisible by 4 and b2. The fac- tors of c5 are c4 and c. 5 is the coefficient of the radical expression. Multiply the coefficient 5 with the 4th root of 16a4 b8 c4. PIVOT 4A CALABARZON Math G9 20 Simplifying Radicals by Reducing the Order of Radicals To reduce the order of radicals is to reduce the index to its lowest possible number. Examples: 1. Simplify 8 a12. Solution: 8 a 12  8 a 8.a 4 ,find a factor of radicand that is a power of 8. = 8 a8  8 a 4 ,find the 8th root of each factor. 8 4 = a a 4 = a  a8 ,change the radical to exponential form. 1 = aa 2 ,reduce the rational exponent to lowest term. = a a ,change the exponential form to radical form. 2. Simplify 4 25. Solution: 25 is not a perfect power of 4 but a perfect power of 2. Hence you can express 25 as 52 4 25  4 52 2 1  54  52 ,change to exponential form and reduce the rational exponent to lowest term. 1 5 2  5 ,change the exponential to radical form. Therefore, 4 25  5 is in simplest form. 3 3. Simplify 64. Solution: Simplify the innermost radical 3 64  3 8 ,the square root of 64 is 8, since 82 = 64. = 2 ,the cube root of 8 is 2, since 23 = 8. 4. Simplify 6 27m 3 n 3. Solution: 6 27m 3 n 3  6 33 m 3 n 3  6 (3mn) 3 ,express 27 as power of 3: 33 , then apply the laws of exponent. 3 1 6 (3mn) 3  (3mn) 6  (3mn) 2 ,change radical to exponential form and reduce the rational exponent 1 to lowest term. (3mn) 2  3mn ,change exponential form to radical form. Therefore, the simplest form of 6 27m n 3 is 3 3mn. 8 5. Simplify 81. 1  814   1 1 81  81 1 1 1 2 Solution: 8 8  4 81  3 , the exponent     8 4 2 21 PIVOT 4A CALABARZON Math G9 Simplifying Radicals by Rationalizing Denominators In a radical expression, if there is radical number in the denominator, the expression is not in simplest form. To eliminate the radical number in the de- nominator you should apply the rationalization process. To rationalize denominator, you have to multiply the numerator and denomi- nator by a number such that the resulting radicand in the denominator is a per- fect nth power. Examples: 3 1. Simplify. 2 Solution: Apply quotient law of radical 3 3  2 2 Express as quotient of two radicals. 3 2 6   Multiply the radical fraction by a radical which will 2 2 4 make the radicand in the denominator a perfect square. 6 = Simplify. 2 Remember that a radical fraction is in its simplest form if there is no radical number in the denominator. 3 48 2. Simplify 3. 6 48 3 Solution: 3  82 Apply the quotient law of radicals then find the 6 quotient of the radicand. The quotient 8 is a perfect cube. Thus, the cube root of 8 is 2. 32 3. Simplify: 3 54 3 32 3 84 Solution:  Simplify the radicand by finding a factor which is a 3 54 3 27  2 perfect cube. 23 4  Extract the cube root of the factors that are perfect 33 2 cube. 23  2 Simplify by dividing the radicands 3 PIVOT 4A CALABARZON Math G9 22  64 3 4. Simplify 4. 2 Solution: Find smallest possible number where you can extract the 4th root. Since the radicand in the denominator is 2, you need to find a number when multiplied by 2 will give you a number where you can find the 4th root. In this case the, radicand in the denominator must be 16. Thus multiply both radicand by 8. 64 3  8 64 24 64 24 Simplify by extracting the 4th root  4   34 24 4 2 8 16 2 of 16, then divide 6 by 2. Conjugate of Radical Expression Examples: Radical Expression Conjugate 3 2 3 2 5 2 5 2 The product of the radical expression and its conjugate is an integer. In finding its product, the same procedure is applied as in, multiplying the sum and difference of 2 binomials like, (x + y)(x– y) = x2 - y2. Finding the product of radical expression and its conjugate: A. (3  2 )(3  2 )  (3) 2  ( 2 ) 2  9  2  7 B. ( 5 2 )( 5  2  52  3 3 4. Rationalize the expression. 2 3 Solution: Find the conjugate of the denominator then multiply the numerator and the denominator by the conjugate. 3(2  3 ) 3(2  3 )   3(2  3 ) Simplify the denominator. (2  3 )(2  3 ) 43 23 PIVOT 4A CALABARZON Math G9  E Learning Task 1 A. Simplify the radical expressions. 1. 63 6. 3 24 11. 4 625 2. 48 7. 3 81 12. 5 96 3. 75 8. 3 128 13. 6 128 4. 99 9. 3 40 14. 4 243 5. 92 10. 3 135 15. 3 3000 B. Simplify radicals by reducing the order or index of radical. 6 1. 16 3. 10 32 5. 12 729 12 2. 3 64 4. 81 A A. Find the product of the radical expression and its conjugate. Radical Expression Conjugate Product 1. 2. 3  5 3. 3 2 B. Simplify 6 3 by rationalizing the denominators. 1. 6. 8 1 2. 2 7. 5 2 3 5 4 3. 4 8. 62 4 5 2 4. 8 9. 7 3 14 5 5. 7 10. 4 5 3 9 2 3 6 2 1 PIVOT 4A CALABARZON Math G9 24  WEEK Operations of Radicals 7 Lesson I In this lesson, you have to perform operations of radicals such as: A. adding and subtracting similar and dissimilar radicals; and B. multiply and/or divide radicals. Radical terms are similar if they have the same radicand and index. Examples: 5 ,3 5 , 5 These three radical numbers are similar since the radicand are the same which is 5 and the index is 2 or all square root. Radicals are unlike or dissimilar if the radicand are not the same and/or the index are also different. Examples: 5 , 3 5 The radicals have the same radicands but the indices are different. 5 , 6 The indices are the same but the radicands are different. D Addition and Subtraction of Radicals In adding or subtracting radicals, you can only add or subtract similar radi- cals. To add or subtract similar radicals, add or subtract only the coefficients of the radical number and copy the common radicals. Examples: 1. 4 2 2  2 2  ( 4  1  2) 2  3 2. Add and/or subtract the numerical coefficients. If there is no number before the radical number it is understood that the coefficient is 1. 2. 5 4  2 4  (5  2) 4  3 4  3( 2)  6. Subtract the numerical coefficients Since the radicand is a perfect square, get the square root and multiply the root by the coefficient. 3. 3 3  5 3  2 5  (3  5) 3  2 5  8 3  2 5. Add only the coefficients of terms that are similar. 4. 2 50  8  3 2. In this case we can simplify first the radicand. Since the lowest radicand is 2 find two factors of the other radicand in such a way that one factor is 2 and the other is a perfect square. 2 25  2  4  2  3 2. The factors of 50 are 25 and 2, where25 is a perfect square and the factors of 8 are 4 and 2, where 4 is also a perfect square. 25 PIVOT 4A CALABARZON Math G9  2(5) 2  2 2  3 2 Simplify by getting the square root of numbers which are perfect square. Now that the radicals are similar, you can combine them. 10 2  2 2  3 2  (10  2  3) 2  9 2 Simplify. 5. Simplify: 2 x3 8  3x  53 27 x 3  3x Simplify the radicand by finding a factor which are perfect cube. 2 x( 2)3 3 x  5(3 x)3 3 x Extract the cube roots. 4 x 3 3 x  15 x 3 3 x  19 x 3 3 x Simplify and combine similar terms. Multiplication of Radicals There are three cases to be considered in multiplying radicals. 1. Multiplying radicals of the same order or index. Apply the rule n a  n b  n ab Examples: A. Find the product of 3 5  2. Solution: 3 5  2  3(1) 5  2 Multiply the coefficients and the radicands.  3 10 B. Find the product of  43 2  33 4. Solution:  43 2  33 4  (4)(3)3 2  4  123 8 Multiply both coefficients and radicands. = (-12)(2) Extract cube root of 8. = -24 C. Find the product of 32 xy 3  18 x 2. Solution: Simplify first the radicand. 32 xy 3  18 x 2  16 y 2  2 xy  9x2  2  4 y 2 xy  3 x 2  4 y (3 x) 2 xy (2) Multiply the coefficients and radicands.  12 xy 4  xy Extract square root of 4.  12 xy ( 2) xy Simplify.  24 xy xy PIVOT 4A CALABARZON Math G9 26 2. Multiplying Radicals with different indices but same radicands. If the radicands are the same but different indices, you have to change them into similar index. Examples: 1. Find the product of 3  3.3. 1 1 Solution: 3  3 3  3 2  33 Change to exponential form. 1 1 = 3  2 3 Apply the law of exponent. 3 2 = 3  6 6 Change the exponential fraction to similar fractions. 5 = 3 6 Add the fractions. = 6 35 Change to radical form. = 6 243 Find the power of 35. 2. What is the product of 3 2  4 4? Solution: 3 2 4 4  3 2.4 2 2 Rename 4 to make the radicands the same. 1 2  2  2  3 4 Change to exponential form. 1 1  2  Reduce 2 to lowest term and apply the law  3 2 4 of exponent. 2 3 5  2 6  2 6  6 Change the exponential fractions to similar fractions, then add.  6 25 Change to radical form  6 32 Find the power of 25. Binomials can be multiplied using distributive property or applying the rule of special products of algebraic expressions or binomials. 3. Find the product and simplify: 3 2 4 2  3  Solution: 3  2 4 2 3 3 2 4  2 3 2  3  Apply distributive property.  3(4) 2  2  3(1) 2  3 Multiply both coefficients and radicands.  12(2)  3 6  24  3 6 Simplify. 4. Find the product of the sum and difference of the radical binomials 2  3 3 2 2 3 3 2.  Solution: Apply the rule in multiplying sum and difference of binomials. 2  3 3 2 2 3 3 2  2 3     3 2  2 2  4(3)  9(2)  12  18  6 27 PIVOT 4A CALABARZON Math G9 Dividing Radicals If the nth root of a number is divided by the nth root of another number, then the result is the nth root of the quotient. n p p , where q≠0  n n q q Dividing radicals with the same indices Examples: 3 16 1. Find the quotient: 2 2 3 Solution: 16 16 Find the quotient of 16 and 2. The quotient  3  3 82 3 2 2 is 8 which is a perfect cube. Hence, the cube root of 8 is 2. 2. Divide 8 by 5. Solution: 8 8 5 Rationalize the denominator by a number   5 5 5 that will make the denominator a perfect square. 40 = 25 Multiply. 4  10 = 25 Factor the numerator of the radicand. 2 10 = 5 Simplify. 3. Find the quotient: 3 6 6 3 Solution: Change first to same index. 1 2 3 6  6 3  6 6 1 1 36 36 6 3 6 2 6 (3  2) 2 6  3 3 Simplify 32  2 2 6 6 3  3  4  6 12 Find the quotient then, simplify. 4. Divide 8 by 3  3. Solution: To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator. 8 3 3 8(3  3 )   3 3 3 3 (3  3)(3  3 ) 8(3  3 ) 8(3  3 ) 4   (3  3 ) = 93 6 3 Multiply the sum and difference of binomial radicals then, sim- PIVOT 4A CALABARZON Math G9 28  E Learning Task 1 A. Make each pair of radicals similar by reducing the radicand. 1. 2, 8 6. 48 , 12 2. 3 3 , 3 24 7. 28 , 63 3. 45 , 125 8. 4 32 , 4 162 3 4. 5 3 , 5 96 9. 40 , 3 135 5. 27 , 75 10. 8 , 200 B. Referring to each pair in letter A, do the following. A. Find the sum of each pair of radicals. B. Subtract the first radical from the second radical. C. Find the product of each pair of radicals D. Divide the second radical by the first radical. A A. 1. What is/are the rule(s) in adding or subtracting radicals? 2. What is/are the rule(s) in multiplying radicals? 3. How do you divide or rationalize radicals? 4. How do you rationalize if the denominator is a binomial radicals. B. Perform the indicated operations. Express your answer in simplest radical form. 1. 11. 9 2. 25 3. 12. 2 4. 4 8 5. 13. 33 2 6. 3 3 7. 14. 5 8. 5 2 9. 15. 10. 3  4 5x 2 4 3x 3 29 PIVOT 4A CALABARZON Math G9 WEEK 8 Radical Equations I Lesson A radical equation is an equation whose unknown quantity is in the radicand. Below are examples of radical equations and not radical equations. Radical Equations Not Radical Equations 1. 3 x  5  20 1. x 3  5  20 2. x  3  2 3 3 3 2. 2x  3  x  2 Some variable x are found Variable x is not part of radi- in the radicand. cand. In this lesson you have to solve radical equations. Hence, you have to assume that if two numbers are equal, then the square, cube or nth power are also n   equal. If x = y, then xn = yn or if n x  y then n x  y n and x = y n. D Solving Radical Equations In solving radical equations, you must do the following steps: 1. Write the equation such that the radical containing the unknown is on one side of the equation. 2. Combine similar terms. 3. Raise both sides of the equation to a power same as the index of the radical. The equation should be free of radical to complete the solution. 4. Check if the value or values obtained will make the original equation true. Examples: 1. Solve for the value of x in x 7 Solution:  x 2  72 Since the index is 2, raise both sides of the equation to power of 2. x = 49 Check: x 7 49  7 7 = 7 Therefore, the square root of 49 is 7. PIVOT 4A CALABARZON Math G9 30  2. Solve: 3 x6 3 Solution:  3 x6  3  33 Cube both sides of the equation since the index of the radical is 3. x - 6 = 27 x = 27 + 6 By APE, combine similar terms. = 33 Check: Substitute the value of x to the original equation 3 33  6  3 3 27  3 3 =3 Therefore, the cube root of 27 is 3. 3. Solve : 4 x2  x Solution: x2  x4 By APE, radical should be in one side of the equation  x2 2  x  4 2 Square both sides of the equation since the index of the radical is 2. x  2  x 2  8 x  16 Square of a binomial. x 2  9 x  18  0 Combine similar terms. x  6( x  3)  0 Solve quadratic equation by factoring. x - 6 = 0 or x - 3 = 0 Use zero product property. x = 6 or x=3 Check:: For x = 6 For x = 3 4 x2  x 4 x2  x 4 62 6 4 32  3 4 4 6 4 13 426 4 1  3 6= 6 5 ≠ 3 Only 6 makes the equation true. Hence, 6 is the solution of the equation. 3 is an extraneous root of the quadratic equation. 31 PIVOT 4A CALABARZON Math G9 4. The square root of the sum of a number and 9 is 6. Find the number. Solution: Let x = the number x + 9 = the sum of the number and 9 x  9  6 Equation  x9  2  6  Square both sides of the equation since the index is 2 2 x + 9 = 36 Simplify. x = 36 - 9 By APE x = 27 Check: x9  6 27  9  6 Replace x by its value. 36  6 6= 6 Common problem involving radical is the solution of right triangle applying the Pythagorean Theorem. This theorem is about the relationships among the sides of a right triangle. The theorem states the “sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. Thus, if a and b are the legs and c is the hypotenuse of the right triangle, then c 2  a 2  b 2 or c a2  b2 Examples: 1. A man walks 8 meters to the east ,then turn south and walk 6 meters. How far is he from his starting point? Solution: Visualize the problem and sketch the figure. Let A be the starting point , B the ending point and C the turning point. The distance of the man from the starting point is the hypotenuse of the right triangle. c a2  b2 c  6 2  82 c  36  64 c 100 c = 10 Therefore, the distance of the man from his starting point is 10 meters. PIVOT 4A CALABARZON Math G9 32 2. A ladder of 12 ft. tall leans against the wall of a building. The ladder touches a point of the wall 6 ft. from the ground. How far is the foot of the ladder from the building? Solution: Sketch the figure. Let l = 12 Length of the ladder which becomes the hypotenuse of the right triangle. h=6 Distance of the top of the ladder from the foot of the building. d=? Distance of the foot of the ladder from the foot of the building. l 2  h2  d 2 d 2  l 2  h2 by APE d  l 2  h2 d  122  62 d  144  36 d  108 d 36  3  6 3 The distance of the foot of the ladder from the foot of the building is 6 3 ft. 3. The area of a square lot is 162 m2. How long is the side of the square? Solution: A = s2 The area of a square. 162  s 2 Substitute the given in the formula. s 162 Solve for s. s  81 2 Simplify the radicand by finding a factor that is a perfect square. s9 2 m The length of the side of the square. 4. The diagonal of the rectangle is 5 x. If the width of the rectangle is x and the length is twice the width, what is the dimension of the rectangle? Solution: Using the Pythagorean theorem c2  a 2  b2 5 x 2  2 x    x  2 2 Since the width is 5 units, then the length is 2(5) = 10 25 x  4 x 2  x 2 5 x units and the diagonal is 5 5 25 x  5 x 2 units. 25 x 5x2  5x 5x x = 5 units 33 PIVOT 4A CALABARZON Math G9  E Learning Task 1: Solve the following equations. 1. x 9 11. 2 x  2  x  10 2. 3 3x  6 12. x 2  144  0 3. 4 x 6 13. x x 2  3  3x 4. 7  x  12 14. 2 3x  1  4 x 5. 4 3x  5  8 15. 2  x 1  x  5  0 6. 3x  2  5 7. 5 x 5 7 8. 3 x  33 7 9. 84 x  7 10-. 33 x  12 A Solve the following problems. 1. Twice the square root of a number is 12. Find the number. 2. The square root of a number increased by 9 is 27. find the number. 3. The square root of the sum of twice a number and 3 is 6. Find the number. 4. The perimeter of the square is 25 and the side is x  3. Find the number. 5. The circumference of the circle( C = 2πr) is 24 cm and the radius is x  2. Find x and its radius. 6. Find the side of the square if the area is 64m2. 7. The shorter l6g of a right triangle is one half of the longer leg and the hypotenuse is 3 cm. How long is each leg of the triangle? 8. Five times the square root of a number is five. Find the number. 9. The cube root of a number decreased by 3 is zero. Find the number. 10. The square root of the cube root of a number is 2. find the number. PIVOT 4A CALABARZON Math G9 34 PIVOT 4A CALABARZON Math G9 35 Week 4 Learning Task 1 Learning Task 2 Learning Task 3 2 5 3 10 m3n 1. n12 A. 1.  125  B. 1. 73 B. 1. m n 6 6. 1 2 2. 3b2 2. 5 x 2. 11x 2 2. q3 3 8 6 4  p 3. a 3.  16  3. 4m 5 7. x6 y3   2 1 5 x 5 5 6 3 4. 432 4.  xy  4. 2a 4 3. 8.  2a    a 1 a2 12 3 4 5 5. 5.a b 5. 5  5 52 5. 84 3 4. b 9. 2x2