LESSON 2: SIMPLIFYING RADICALS

Transcription

1 High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN N.RN. 4. N.RN. 5. N.RN N.RN. 7. A 7 N.RN Copyright 05 Pearso Educatio, Ic.

2 High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS Challege Problem N.RN. 9. If the two radical epressios are equivalet, you have: 6 64 This meas that: 6 64 ( ) ( ) You ca compare the terms iside the radicals to figure out the value of. The value of should solve both equatios: The equatios are equivalet whe 5. Therefore the two radical epressios are ad. Copyright 05 Pearso Educatio, Ic. 4

3 High School: Workig with Epressios LESSON : NUMBER SYSTEM N.RN.. 5 or 5 or 5 or 5 N.RN.. N.RN.. N.RN or or 5+ or or 5. or 5. or 5. or. 5 or. 5 or. 5 or. 5 or. 5 or. 5 N.RN. 5. B 5 C.. E 4 9 N.RN. 6. Ratioal Irratioal Ca't Tell N.RN. 7. B a irratioal Copyright 05 Pearso Educatio, Ic. 5

4 High School: Workig with Epressios LESSON : NUMBER SYSTEM Challege Problem N.RN. 8. A ratioal umber is a umber that ca be represeted as a fractio of two itegers. If umber is a repeatig umber, you ca always covert this umber ito a fractio of two itegers. To fid those itegers, use epoets of 0 to create a subtractio operatio usig two umbers derived from that both oly have the repeatig digits i their decimal portio. For eample, take the repeatig decimal umber The repeatig portio of this umber is 87. Create two umbers that oly have the repeatig portio as their decimal portio by multiplyig by epoets of 0 (e.g. 05 ad 0). Subtractig these two umbers will get rid of the decimal portio. The outcome of the subtractio will be a iteger. 5 0,58, ,58,406 (0 5 0),58,406,58,406,58,406,69, ,990 49,9 95,69,0 So, ,995 As with this eample, all repeatig decimals ca be coverted to fractios. Copyright 05 Pearso Educatio, Ic. 6

5 High School: Workig with Epressios LESSON 4: POLYNOMIALS A.APR.. A.APR.. A.APR.. A.APR. 4. D Distributive property C Commutative property A Associative property D Distributive property A.APR (9 4) A.APR (7 + ) + (9 ) A.APR. 7. A B ( y) (7 + y + 4 5y ) 4 + ( 7 7) y + (9 4 ) + ( y + 5y ) 4 4 y y y + 5 y A.APR. 8. 9( + 4y ) 7(y ) 9 + 6y 7y y A.APR (9y + 4) (7y ) 4.5y + 4y y 4y + 8 A.APR. 0. To make the operatio simpler, you ca first simplify each polyomial before addig ad subtractig. A (9y y + 4 ) (7y + 4 ) 4y + 8 B 4( + 7y ) (5 + 9y ) 8 + 8y 5 9y (8 5 ) + ( 9y + 8y ) + 9y C (9y + 4 7y ) ( + y ) 8y + 8 4y y (8 ) + (8y 4y y ) 5 + y You ca the fid the polyomial A B + C usig the simplified epressio. A B + C (4y + 8 ) ( + 9y ) + (5 + y ) (8 + 5) + (4 9 + )y 0 4y Copyright 05 Pearso Educatio, Ic. 7

6 High School: Workig with Epressios LESSON 4: POLYNOMIALS A.APR.. 5( ) 7( ) + ( ) (5 7 + )( ) ( ) A.APR.. 9( 4y ) 7 ( 4y + ) + ( y ) 9( 4y ) 7( 4y ) + 4y (9 7 + )( 4y ) ( 4y ) 6 y A.APR.. (7y 4) + 7(7y 4) 9 (7y 4) ( + 7 9)(7y 4) 0 0(7y 4) 0 A.APR. 4. 4( 7y) 9( 7y) ( + 7y) 4( 7y) 9( 7y) + ( 7y) (4 9 + )( 7y) ( 7y) 9 + y y 9 Challege Problem A.APR. 5. Let s call A the missig epressios. The sum of the two epressios is 5. Therefore you ca write: ( 7) + A 5 A 5 ( 7) A + ( + 5) + 7 A The missig epressios is Copyright 05 Pearso Educatio, Ic. 8

7 High School: Workig with Epressios LESSON 5: MULTIPLYING POLYNOMIALS A.APR.. A ( ) C ( + )( + ) E 4 ( + ) + A.APR.. ( + )(4 4) ( )( ) ( )( + )( ) ( 5) ( )( + ) ( ) A.APR A.APR A.APR A.APR A.APR A.APR A.APR A.APR A.APR Copyright 05 Pearso Educatio, Ic. 9

8 High School: Workig with Epressios LESSON 5: MULTIPLYING POLYNOMIALS Challege Problem A.APR. A.REI.. If you multiply the two biomials, you will fid: (a + b)(c + d) ac + (ad + bc) + bd Therefore: R ac S ad + bc T bd Sice a, c, S, ad T 0, you have: R ()() ()d + b() b + d 0 bd From this you ca deduce that R. The easiest way to approimate the values of b ad d is to graph the two equatios remaiig ad fid the poit of itersectio of the correspodig graphs. Sice graphig oly provides a approimatio of the itersectio, you will eed to verify that the coordiates obtaied solve both equatios. If you chage the variables i the equatios for b ad d with ad y, respectively, you obtai: y y 0 There are two possible poits of itersectios: ( 0, ) ad ( 0.5, 0). Both coordiate pairs solve the equatios ad are therefore correct values. As a result, there are two possible sets of biomials that solve the problem: ( 0)( ) + 0 ( 0.5)( 0) + 0 Copyright 05 Pearso Educatio, Ic. 40

9 High School: Workig with Epressios LESSON 6: FACTORING A.SSE.. + b + c + b c b c b + c c ( + A)( + B) ( A)( + B) A < B ( A)( + B) A > B ( A)( B) ( A)( + B) A B A.SSE..a. B ( )( + 4) A.SSE..a ( + )( + 4) ( + )( + 9) ( + 7)( + 8) ( + 4)( 6) ( )( 5) A.SSE..a ( + )( + ) ( + )( 7) ( )( 5) ( + 8)( + 9) ( )(4 5) A.SSE..a 5. ( + 6) ad ( + 9) A.SSE..a 6. ( + 4) ad ( + ) A.SSE..a 7. ( )( 5) A.SSE..a 8. ( + )( + 5) A.SSE..a 9. ( + )( + 9) A.SSE..a 0. ( + 4)( 6) A.SSE..a. ( + )( 5 7) A.SSE..a. 5( + )( ) Copyright 05 Pearso Educatio, Ic. 4

10 High School: Workig with Epressios LESSON 6: FACTORING Challege Problem A.SSE.. You ca multiply the biomial epressios ito a triomial form. (m + p)( + q) m + (p + q) + pq Sice a + b + c (m + p)( + q), you write the followig relatioships: a m b p + q c pq From these relatioships you ca determie the sig of each coefficiet m,, p, ad q depedig o the sigs of a, b, ad c. a. a is a positive umber Sice a m, whe a > 0, the m ad are both either positive or egative. a is a egative umber Sice a m, whe a > 0, the m ad have opposite sigs (m is egative if is positive ad vice versa). b. b ad c are positive umbers b mp + q > 0 c pq > 0 For both equatios to be true, both p ad q must be positive umbers. Both m ad must also be positive umbers. b is a positive umber ad c is a egative umber b mp + q > 0 c pq < 0 For both equatios to be true, p ad q must have opposite sigs ad the absolute value of the positive umber must be greater tha the absolute value of the egative umber. Both m ad must also be positive umbers. p < 0 ad q > 0 ad q > p or p > 0 ad q < 0 ad p > q b is a egative umber ad c is a positive umber b mp + q > 0 c pq > 0 For both equatios to be true, both p ad q must be egative umbers. Both m ad must also be positive umbers. b ad c are egative umbers. b mq + p < 0 c pq < 0 For both equatios to be true, p ad q must have opposite sigs ad the absolute value of the egative umber must be greater tha the absolute value of the positive umber. Both m ad must also be positive umbers. p < 0 ad q > 0 ad q < p or p > 0 ad q < 0 ad p < q Copyright 05 Pearso Educatio, Ic. 4

11 High School: Workig with Epressios LESSON 7: SPECIAL BINOMIALS A.SSE.. B square of a sum A.SSE.. C square of a differece A.SSE.. A differece of two squares A.SSE. 4. This epressio is a differece of two squares, a b, where: a + ad b This special product of biomials ca also be writte i factor form as: (a + b)(a b). Therefore the epressio ca be factored: ( + + )( + ) ( + 6) A.SSE..a 5. Square of a Sum Square of a Differece Differece of Two Squares ( + ) (4 + )(4 ) ( ) 9 4 A.SSE..a 6. This epressio is the square of a sum (a + b), where: a 4 ad b Therefore the polyomial form of this epressio is: a + ab + b : A.SSE. 7. This epressio is the square of a differece (a b), where: a ad b Therefore the polyomial form of this epressio is: a ab + b A.SSE..a 8. This polyomial is the differece of two squares a b, where: a 4 ad b Therefore the polyomial ca be factored to: (a + b)(a b) (4 + )(4 ) Copyright 05 Pearso Educatio, Ic. 4

12 High School: Workig with Epressios LESSON 7: SPECIAL BINOMIALS A.SSE..a 9. This polyomial is i the form a + ab + b, which is the polyomial form associated with the square of a sum (a + b). I this particular epressio you ca have two possible sets of values for a ad b. a 64 ad b 8 As a result: a ±8 ad b ±9 Sice ab >0, it meas that a ad b are either both positive or both egative umbers. Therefore the values of a ad b ca either be: a 8 ad b 9 or a 8 ad b 9 The followig two sets of factors are therefore valid: (8 + 9) or ( 8 9) A.SSE. 0. Miki cofused the sum of squares with the differece of two squares. The product of ad is egative, so the 4 must be subtracted from the square of whe multiplyig. ( + )( ) (+)(-) Challege Problem A.SSE.. The epressio ca be rearraged to brig forward the polyomial equivalet to the square of a differece a ab + b : (a ab + b ) 4a b The first portio of the equatio ca the be factored as the square of the differece of a b: (a b) 4a b Sice 4a b (ab), the epressio ca the be rewritte: (a b) (ab) This epressio is the sum of two squares, which ca be factored: (a b ab)(a b + ab) Copyright 05 Pearso Educatio, Ic. 44

13 High School: Workig with Epressios LESSON 8: DIVIDING POLYNOMIALS A.APR. A.APR A.APR.. B 5 + A.APR. A.APR.6. B A.APR. A.APR A.APR. 5. The polyomial ca be factored to ( + )( + 8). Therefore, the epressio ca be simplified to: ( + )( + 8) The quotiet of the epressio is ( + ). A.APR. A.APR.6 6. The polyomial 0 ca be factored to ( + 4)( 5). Therefore, the epressio ca be simplified to: 0 ( + 4)( 5) The quotiet of the epressio is ( 5). A.APR ( ) 6 ( ) Copyright 05 Pearso Educatio, Ic. 45

14 High School: Workig with Epressios LESSON 8: DIVIDING POLYNOMIALS A.APR. A.APR.6 8. You ca first simplify the epressio by dividig the umerator by : + 8 The resultig polyomial + 8 ca be factored to ( )( + 6). Therefore, the epressio ca be simplified to: + 8 ( )( + 6) + 6 The quotiet of the epressio is ( + 6). A.APR. A.APR.6 9. B 0 5 A.APR. A.APR.6 0. You must fid the divisor A that solves this equatio: A This equatio ca be rewritte: 6 6 A + 4 ( 8 6) + 4 ( + 4)( 4) + 4 ( 4) 8 Therefore, the divisor of this equatio is ( 8) Copyright 05 Pearso Educatio, Ic. 46

15 High School: Workig with Epressios LESSON 8: DIVIDING POLYNOMIALS Challege Problem A.APR. A.APR.6. This equatio ca be rewritte as: m ( + p)( + q) The right factored epressio ca be epressed as a polyomial: m ( + (p + q) + pq) + + (p + q) + + pq Sice the greatest epoet of the left epressio is, the greatest epoet of the right epressio must also be, therefore ( + ) ad. m (p + q) + pq From this equatio, these equalities ca be deduced: m p + q 5 pq 4 If p + q 5, the p 5 q ad (5 q)q 4 (5 q)q q + 5q 4 The possible values of q solvig this equatio are either 7 or. p 5 q If q 7 the p. If q the p 7. Sice p > q, the oly possible combiatio is p 7 ad q. The polyomial operatio with all the coefficiets replaced by their actual values is therefore: ( + 7) Copyright 05 Pearso Educatio, Ic. 47

16 High School: Workig with Epressios LESSON 9: OPERATIONS WITH RADICALS N.RN.. B N.RN N.RN.. N.RN N.RN. 5. N.RN ( + ) 5( 6 ) 0 ( ) ( ) ( 4) 6 N.RN. 7. 7( 7 + 5) N.RN ( ) + + Copyright 05 Pearso Educatio, Ic. 48

17 High School: Workig with Epressios LESSON 9: OPERATIONS WITH RADICALS N.RN ( 5) 4 5 N.RN. 0. C Challege Problem N.RN. A.SSE.. Usig the properties of epoets, you ca rewrite the epressio i a form that match ab : a b ma b b ( mab ) b ( mab ) For this epressio to be equal to ab, you must fulfill these two coditios: ( m) ad Therefore: m Sice b b., if, the b b b Therefore whe m ad are replaced i the epressio by their respective values, the epressio becomes: ab ( ) ab b This epressio is ideed equal to ab whe it is simplified. Copyright 05 Pearso Educatio, Ic. 49

18 High School: Workig with Epressios LESSON 0: SOLVING RADICAL EQUATIONS A.REI.. B 6 9 A.REI.. A A.REI.. A B C A.REI You eed to substitute ( ) 6 for i the equatio to check your aswer. This aswer correctly solves the equatio. 6 Copyright 05 Pearso Educatio, Ic. 50

19 High School: Workig with Epressios LESSON 0: SOLVING RADICAL EQUATIONS A.REI ( + ) ( 7 ) You eed to substitute 5 for i the equatio to check your aswer. You also eed to remember that square roots ca oly be used o positive umbers. The square root of a egative real umber is impossible to calculate. ( 5) + ( 5) 7 This solutio is etraeous ad should ot be icluded i your aswer. Sice 5 was the oly possible solutio, this equatio caot be solved. There are o solutios to this problem. A.REI ( 7 ) You eed to substitute 6 for i the equatio to check your aswer This aswer correctly solves the equatio. 6 Copyright 05 Pearso Educatio, Ic. 5

20 High School: Workig with Epressios LESSON 0: SOLVING RADICAL EQUATIONS A.REI ( + ) ( ) You eed to substitute 576 for i the equatio to check your aswer This aswer correctly solves the equatio. 576 A.REI ( ) You eed to substitute ( ) 4 for i the equatio to check your aswer. This aswer correctly solves the equatio. 7 Copyright 05 Pearso Educatio, Ic. 5

21 High School: Workig with Epressios LESSON 0: SOLVING RADICAL EQUATIONS A.REI. 9. ( ) ( ) You eed to substitute 0 for i the equatio to check your aswer ( ) Sice you caot divide by 0, this solutio is etraeous ad should ot be icluded i your aswer. Sice this was the oly possible solutio, this equatio caot be solved. There are o solutios to this problem. Challege Problem A.REI. 0. Square roots of real umbers ca oly be calculated if the umber is positive. Square roots of egative umbers are impossible. The distace betwee two poits is epressed as the square root of the sum of the squares of the differeces betwee the coordiates ad the y coordiates of the two poits. Sice the differeces are squared, the square of the differeces will always be positive umbers eve whe the differeces themselves are egative. Sice both squares of the differeces are positive, the sum of those squares will also be positive. Therefore the umber iside the square root will always be positive ad a distace ca always be calculated irrespectively of the coordiates of the poits. Copyright 05 Pearso Educatio, Ic. 5

22 High School: Workig with Epressios LESSON : PUTTING IT TOGETHER 8.EE.5 8.EE.6. Defiitios ad eamples will vary. Here are some eamples. Cocept or Property egative iteger epoet Descriptio Eamples For ay ozero umber a ad a positive iteger : a 8 a 0 as a epoet For ay ozero umber a: a ,900 0 fractio as a epoet properties of epoets For a > 0: a a m,, ad p are positive itegers: a m + a m a a m a a m (a m ) a m a b m p a b mp p 4 + (4 4) (4 4 4) (4 ) (4 4 4) (4 4 4) (4 4 4) ratioal umbers Numbers that ca be epressio as a ratio betwee two itegers such that the secod iteger is ot zero. All itegers are icluded i ratioal umbers sice they are all divisible by. All decimals that termiate are ratioal sice they are divisible by a factor of 0. All decimals that repeat after some poit are ratioal. All roots of perfect umbers are ratioal umbers (because they are itegers). The system of ratioal umbers is closed uder additio, subtractio, multiplicatio, ad divisio., 0.56, ad 57. are ratioal umbers because they ca all be represeted by a fractio: , Copyright 05 Pearso Educatio, Ic. 54

23 High School: Workig with Epressios LESSON : PUTTING IT TOGETHER 8.EE.5 8.EE.6. Cocept or Property irratioal umbers Descriptio Numbers that caot be writte as a fractio (a ratio of two itegers). I decimal form, irratioal umbers ever ed or repeat. All square roots of umbers that are ot perfect squares are irratioal. Special umbers such as π are also irratioal. real umbers A set of umbers cotaiig all ratioal umbers ad all irratioal umbers. polyomial epressio factorig secoddegree polyomials All the umbers o the umber lie are real umbers. A epressio of oe or more algebraic terms with whole-umber (positive itegers) epoets. Sice the epoet has to be a positive iteger, epressio cotaiig iverse of epoets or roots are ot polyomial epressios. Secod-degree polyomials ca sometimes be factored ito two biomials with coefficiets. The epressio a + b + c ca be factored ito (m + p)( + q), where m a, pq c, ad (p + mq) b. Eamples 445. π polyomial epressios: + + ot polyomial epressios: ( + )( 5) Copyright 05 Pearso Educatio, Ic. 55

24 High School: Workig with Epressios LESSON : PUTTING IT TOGETHER 8.EE.5 8.EE.6. Cocept or Property special products of biomials Descriptio Eamples There are three special products of biomials: The square of a sum: (a + b) (a + b)(a + b) a + ab + b The square of a differece: (a b) (a b)(a b) a ab + b The differece of two squares: (a + b) (a b) a b The square of a sum: ( + ) + ()() + (5) The square of a differece: ( ) ()() + ( ) The differece of two squares: ( + )( ) (5)( ) Copyright 05 Pearso Educatio, Ic. 56