Summer MA Lesson 13 Section 1.6, Section 1.7 (part 1)

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Suer MA 1500 Lesso 1 Sectio 1.6, Sectio 1.7 (part 1) I Solvig Polyoial Equatios Liear equatio ad quadratic equatios of 1 variable are specific types of polyoial equatios. Soe polyoial equatios of a higher degree ca be solved by factorig. 1. Always look for a GCF first.. Factor usig a differece of squares, trioial ethods, or groupig. Ex 1: 6y 1y y 4 Ex : 4 4x 97x 0 Ex : x x x 6 108 0 II Solvig radical equatios Just as both sides of a equatio ay have the sae uber added or be ultiplied by the sae uber, both sides of a equatio ay be raised to the sae power. However, we ust be careful. Power Property for equatios: Whe both sides of a equatio are raised to the sae power, the solutios that result ay be solutios of the origial equatio. For exaple: x solutio is x 4 solutios are ad 1

Raisig both sides to the sae power ay result i a equatio ot equivalet to the origial equatio (differet solutio sets). These extra solutio or solutios (such as the - solutio for the secod equatio) is/are kow as extraeous solutio(s) (solutios that do ot satisfy the origial equatio). Therefore, wheever the power property for equatios is used, all possible solutios ust be checked i the origial equatio. Solvig Radical Equatios: 1. Isolate the radical expressio o oe side of the equatio or put oe radical o each side.. Raise both sides of the equatio to a power equal to the idex.. Solve the result. 4. Reeber to check all possible solutios i the origial equatio, sice the power property for equatios was used. Hit: If a bioial is o the side opposite the radical side, FOIL ust be used whe squarig. Ex 4: x 1 Ex 5: x16 5 x Ex 6: 5 x x 1 Ex 7: 1 x8x 1 0

III Solvig Basic Equatios with Ratioal expoets x 1 x x This is the idea whe solvig with ratioal expoets. Raise both sides of the equatio to the reciprocal power so that the variable expressio is to the first power. Solvig equatios of the for x k If is eve: x k x k If is odd: x k x k Ex 8: Solve each. Reeber to check. a) x 16 b) x1 7 IV Solvig Absolute Value Equatios If x, what value(s) could x equal? You should see there could be values for x, - or. Most absolute value equatios will have two solutios, such as this equatio. This is because a absolute value of a egative value or positive value would both be positive. Reeber, absolute value eas distace fro zero ad distace is always positive. Exceptios will be equatios such as x 0 or x 5. If a absolute value equals zero, there is oly oe value of x that will give zero, sice oly the absolute value of 0 is 0. A equatio of the for absolute value equal a egative will ever be true. This type of equatio will always be o solutio, icosistet.

Solvig Absolute Value Equatios: x k 1. If k > 0, the the equatio becoes two liear equatios x k ad x k.. If k = 0, the equatio becoes the liear equatio x = 0.. If k < 0, there is o solutio. Ex 9: x 5 4 Ex 10: 6 x 5 15 Hit: Isolate the absolute value. Ex 11: x 8 Ex 1: x 1 4

V Forulas Ex 1: The forula M 0.7 x1.5 represets the average uber of o-progra iutes i a hour of prie-tie cable x years after 1996. Project whe there will be 16 iutes of o-progra iutes of every hour of prie tie cable TV, if this tred cotiues. Ex 14: For each plaet i our solar syste, its year is the tie it takes the plaet to revolve oce aroud the su. The forula E 0.x odels the uber of Earth days i a plaet s year, where x is the distace of the plaet fro the su, i illios of kiloeters. Assue a plaet has a orbit equal to 00 earth days. Approxiate the uber of kiloeters that plaet is fro the su. VI Represetig a Iequality There are ways to represet a iequality. (1) Usig the iequality sybol (soetie withi set-builder otatio), () usig iterval otatio, ad () usig a uber lie graph. The followig table illustrates all three ways. Notice that iterval otatio looks like a ordered pair, soeties with brackets. Whe writig the ordered pair, always write the lesser value to the left of the greater value. A parethesis ext to a uber illustrates that x gets very, very close to that uber, but ever equals the uber. A bracket ext to a uber eas it ca equal that uber. With or a parethesis is always used, sice there is ot a exact uber equal to or. 5

Table 1.4 o page 174 of the textbook also illustrates the ways to represet a iequality where a < b. Ex 15: Write this iequality i iterval otatio ad graph o a uber lie. {x x > 1} Ex 16: Write the set of ubers represeted o this uber lie as a iequality ad i iterval otatio. ( ].5 5.7 Ex 17: Write the followig as a iequality ad graph o a uber lie. (,5] Exaie the uber lie below. -10-8 ] ( -6-4 - 0 4 6 10 8 I set-builder otatio, it would be represeted as { x x 6 or x 1} ad i iterval otatio, it would be represeted as (, 6] (1, ). 6

VII Solvig a Liear Iequality i Oe Variable Solvig liear iequalities is siilar to solvig liear equatio, with oe exceptio. Exaie the followig. 5 10 Add 6 to both sides: 1 4 True Multiply both sides by : 10 0 True Divide both by 5: 1 True However, try ultiplyig by : 10 0 False Divide by 5: 1 False. This leads to the followig properties of Iequalities 1) The Additio Property of Iequality If a b, the a + c b + c a- c b- c ) The Positive Multiplicatio Property of Iequality If a b ad c is positive, the ac < bc a b c c ) The Negative Multiplicatio Property of Iequality If a b ad c is egative, the ac > bc a b c c Ex 18: Solve each iequality. Write the solutios with the iequality sybol, i iterval otatio, ad graph the solutios o a uber lie. 7

a) ( x ) 4(5 x) b) 6( x4) ( x) 5 4 5 4 c) ( a ) a ( a ) 1 9 It the variables drop out of a liear iequality, the solutio is either all real ubers (except for ay that ay ot be i the doai) or there is o solutio. ( x ) x 7 x 6 x 7 6 7 The result above is always true, 6 is less tha 7. The solutio is { x x is a real uber} or or (, ). ( x ) x x 6 x 6 Six is ever less tha. The result is false. The solutio is or o solutio. 8

III Solvig Copoud Iequalities Whe solvig a iequality such as 1 x, the goal is to isolate the x i the iddle. Such a iequality is called a copoud iequality ad eas the sae as 1 x ad x. The solutio will be the ubers that, whe substituted i x +, yields betwee 1 ad. 4 x 1 5 Begi by addig 1 to the left, iddle, ad right. Exaple: x 6 Divide the left, iddle, ad right by. x Ay uber betwee -1 ½ ad akes the iequality stateet true. Ex 19: Solve each copoud iequality. Write the solutios usig the iequality sybols, i iterval otatio, ad graph the solutios o a uber lie. a) 1 x 1 b) 6 x 9