Algebra review-olga lednichenko math turtoring, math,
Upcoming SlideShare
Loading in...5
×
 

Algebra review-olga lednichenko math turtoring, math,

on

  • 431 views

OLGA LEDNICHENKO MATH TUTORING, MATH, MATHEMATICS, COLLEGE ALGEBRA, EDUCATION, ELEMENTRAY CALCULUS,

OLGA LEDNICHENKO MATH TUTORING, MATH, MATHEMATICS, COLLEGE ALGEBRA, EDUCATION, ELEMENTRAY CALCULUS,

Statistics

Views

Total Views
431
Views on SlideShare
431
Embed Views
0

Actions

Likes
0
Downloads
3
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Algebra review-olga lednichenko math turtoring, math, Algebra review-olga lednichenko math turtoring, math, Document Transcript

    • Review of Algebra
    • 2 s REVIEW OF ALGEBRA Review of Algebra q q q q q q q q q q q q q q q Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus. Arithmetic Operations The real numbers have the following properties: a b b a ab ba (Commutative Law) a b c a b c ab c a bc (Associative Law) ab c ab ac (Distributive law) In particular, putting a 1 in the Distributive Law, we get b c 1 b c 1b 1c and so b c b c EXAMPLE 1 (a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 3x If we use the Distributive Law three times, we get a b c d a bc a bd ac bc ad bd This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a b c d In the case where c a and d b, we have 2 a b a2 ba ab b2 or 2 1 a b a2 2ab b2 Similarly, we obtain 2 2 a b a2 2ab b2
    • REVIEW OF ALGEBRA x 3 EXAMPLE 2 (a) 2x 1 3x 5 6x 2 3x 10x 5 6x 2 7x 5 (b) x 6 2 x 2 12x 36 (c) 3 x 1 4x 3 2x 6 3 4x 2 x 3 2x 12 12x 2 3x 9 2x 12 12x 2 5x 21 Fractions To add two fractions with the same denominator, we use the Distributive Law: a c 1 1 1 a c a c a c b b b b b b Thus, it is true that a c a c b b b But remember to avoid the following common error: a a a| b c b c (For instance, take a b c 1 to see the error.) To add two fractions with different denominators, we use a common denominator: a c ad bc b d bd We multiply such fractions as follows: a c ac b d bd In particular, it is true that a a a b b b To divide two fractions, we invert and multiply: a b a d ad c b c bc d
    • 4 s REVIEW OF ALGEBRA EXAMPLE 3 x 3 x 3 3 (a) 1 x x x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 x x y 1 y y x y x xx y x 2 xy (d) y x y y x y yx y xy y 2 1 x x Factoring We have used the Distributive Law to expand certain algebraic expressions. We some- times need to reverse this process (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest situation occurs when the expres- sion has a common factor as follows: Expanding 3x(x-2)=3x@-6x Factoring To factor a quadratic of the form x 2 bx c we note that x r x s x2 r sx rs so we need to choose numbers r and s so that r s b and rs c. EXAMPLE 4 Factor x 2 5x 24. SOLUTION The two integers that add to give 5 and multiply to give 24 are 3 and 8. Therefore x2 5x 24 x 3 x 8 EXAMPLE 5 Factor 2x 2 7x 4. SOLUTION Even though the coefficient of x 2 is not 1, we can still look for factors of the form 2x r and x s, where rs 4. Experimentation reveals that 2x 2 7x 4 2x 1 x 4 Some special quadratics can be factored by using Equations 1 or 2 (from right to left) or by using the formula for a difference of squares: 3 a2 b2 a b a b
    • REVIEW OF ALGEBRA x 5The analogous formula for a difference of cubes is 4 a3 b3 a b a2 ab b2which you can verify by expanding the right side. For a sum of cubes we have 5 a3 b3 a b a2 ab b2EXAMPLE 6(a) x 2 6x 9 x 32 (Equation 2; a x, b 3) 2(b) 4x 25 2x 5 2x 5 (Equation 3; a 2x, b 5)(c) x 3 8 x 2 x 2 2x 4 (Equation 5; a x, b 2) x2 16EXAMPLE 7 Simplify 2 . x 2x 8SOLUTION Factoring numerator and denominator, we have x2 16 x 4 x 4 x 4 2 x 2x 8 x 4 x 2 x 2 To factor polynomials of degree 3 or more, we sometimes use the following fact. 6 The Factor Theorem If P is a polynomial and P b 0, then x b is a factor of P x .EXAMPLE 8 Factor x 3 3x 2 10x 24. 3 2SOLUTION Let P x x 3x 10x 24. If P b 0, where b is an integer, thenb is a factor of 24. Thus, the possibilities for b are 1, 2, 3, 4, 6, 8, 12,and 24. We find that P 1 12, P 1 30, P 2 0. By the Factor Theorem,x 2 is a factor. Instead of substituting further, we use long division as follows: x2 x 12 x 2 x3 3x 2 10x 24 x3 2x 2 x2 10x x2 2x 12x 24 12x 24Therefore x3 3x 2 10x 24 x 2 x2 x 12 x 2 x 3 x 4 Completing the SquareCompleting the square is a useful technique for graphing parabolas or integratingrational functions. Completing the square means rewriting a quadratic ax 2 bx c
    • 6 s REVIEW OF ALGEBRA in the form a x p 2 q and can be accomplished by: 1. Factoring the number a from the terms involving x. 2. Adding and subtracting the square of half the coefficient of x. In general, we have b ax 2 bx c a x2 x c a 2 2 b b b a x2 x c a 2a 2a 2 b b2 a x c 2a 4a EXAMPLE 9 Rewrite x 2 x 1 by completing the square. 1 SOLUTION The square of half the coefficient of x is 4. Thus 1 2 x2 x 1 x2 x 1 4 1 4 1 (x 2) 3 4 EXAMPLE 10 2x 2 12x 11 2 x2 6x 11 2 x2 6x 9 9 11 2 2 2 x 3 9 11 2x 3 7 Quadratic Formula By completing the square as above we can obtain the following formula for the roots of a quadratic equation. 2 7 The Quadratic Formula The roots of the quadratic equation ax bx c 0 are b sb 2 4ac x 2a EXAMPLE 11 Solve the equation 5x 2 3x 3 0. SOLUTION With a 5, b 3, c 3, the quadratic formula gives the solutions 3 s32 4 5 3 3 s69 x 25 10 The quantity b 2 4ac that appears in the quadratic formula is called the discriminant. There are three possibilities: 1. If b 2 4ac 0, the equation has two real roots. 2. If b 2 4ac 0, the roots are equal. 3. If b 2 4ac 0, the equation has no real root. (The roots are complex.)
    • REVIEW OF ALGEBRA x 7 These three cases correspond to the fact that the number of times the parabola y ax 2 bx c crosses the x-axis is 2, 1, or 0 (see Figure 1). In case (3) the quad- ratic ax 2 bx c can’t be factored and is called irreducible. y y y 0 x 0 x 0 x FIGURE 1Possible graphs of y=ax@+bx+c (a) b@-4ac>0 (b) b@-4ac=0 (c) b@-4ac<0 EXAMPLE 12 The quadratic x 2 x 2 is irreducible because its discriminant is negative: b2 4ac 12 41 2 7 0 Therefore, it is impossible to factor x 2 x 2. The Binomial Theorem Recall the binomial expression from Equation 1: 2 a b a2 2ab b2 If we multiply both sides by a b and simplify, we get the binomial expansion 3 8 a b a3 3a 2b 3ab 2 b3 Repeating this procedure, we get 4 a b a4 4a 3b 6a 2b 2 4ab 3 b4 In general, we have the following formula. 9 The Binomial Theorem If k is a positive integer, then k kk 1 k 2 2 a b ak ka k 1b a b 1 2 kk 1 k 2 k 3 3 a b 1 2 3 kk 1 k n 1 a k nb n 1 2 3 n kab k 1 bk
    • 8 s REVIEW OF ALGEBRA EXAMPLE 13 Expand x 2 5. SOLUTION Using the Binomial Theorem with a x, b 2, k 5, we have 5 5 4 3 5 4 3 2 x 2 x5 5x 4 2 x 2 2 x 2 3 5x 2 4 2 5 1 2 1 2 3 x5 10x 4 40x 3 80x 2 80x 32 Radicals The most commonly occurring radicals are square roots. The symbol s1 means “the positive square root of.” Thus x sa means x2 a and x 0 Since a x 2 0, the symbol sa makes sense only when a 0. Here are two rules for working with square roots: a sa 10 sab sa sb b sb However, there is no similar rule for the square root of a sum. In fact, you should remember to avoid the following common error: | sa b sa sb (For instance, take a 9 and b 16 to see the error.) EXAMPLE 14 s18 18 (a) s9 3 s2 2 (b) sx 2 y sx 2 sy x sy Notice that sx 2 x because s1 indicates the positive square root. (See Appendix A.) In general, if n is a positive integer, x n sa means xn a If n is even, then a 0 and x 0. 3 3 4 6 Thus s 8 2 because 2 8, but s 8 and s 8 are not defined. The fol- lowing rules are valid: n n n n n a sa sab sa sb n b sb 3 3 3 3 3 EXAMPLE 15 sx 4 sx 3x sx 3 sx xsx
    • REVIEW OF ALGEBRA x 9 To rationalize a numerator or denominator that contains an expression such assa sb, we multiply both the numerator and the denominator by the conjugate rad-ical sa sb. Then we can take advantage of the formula for a difference of squares: (sa sb )(sa sb ) (sa )2 (sb )2 a b sx 4 2EXAMPLE 16 Rationalize the numerator in the expression . xSOLUTION We multiply the numerator and the denominator by the conjugate radicalsx 4 2: sx 4 2 sx 4 2 sx 4 2 x 4 4 x x sx 4 2 x (sx 4 2) x 1 x (sx 4 2) sx 4 2 ExponentsLet a be any positive number and let n be a positive integer. Then, by definition, 1. a n a a a n factors 0 2. a 1 n 1 3. a an 4. a1 n n sa am n n sa m (sa )m n m is any integer 11 Laws of Exponents Let a and b be positive numbers and let r and s be any rational numbers (that is, ratios of integers). Then ar s 1. a r as ar s 2. ar s 3. a r a rs as r r a ar 4. ab a rb r 5. b 0 b brIn words, these five laws can be stated as follows: 1. To multiply two powers of the same number, we add the exponents. 2. To divide two powers of the same number, we subtract the exponents. 3. To raise a power to a new power, we multiply the exponents. 4. To raise a product to a power, we raise each factor to the power. 5. To raise a quotient to a power, we raise both numerator and denominator to the power.
    • 10 s REVIEW OF ALGEBRA EXAMPLE 17 (a) 28 82 28 23 2 28 26 214 1 1 y2 x2 2 2 x y x2 y 2 x 2y 2 y2 x2 xy (b) 1 1 x y 1 1 y x x 2y 2 y x x y xy y x y x y x xy y x xy (c) 43 2 s43 s64 8 Alternative solution: 43 2 (s4 )3 23 8 1 1 (d) 3 4 x 4 3 sx x4 3 3 4 x y 2x x3 y 8x 4 (e) x 7y 5z 4 y z y3 z4 Exercises q q q q q q q q q q q q q q q q q q q q q q q q q q A Click here for answers. 1 1 c 1 11–16 s Expand and simplify. 27. 28. 1 1 1 1 1 1. 6ab 0.5ac 2. 2x 2 y xy 4 c 1 1 x 3. 2x x 4. 4 s s s s s s s s s s s s s s s s s s s s 5 3x x 5. 24 3a 6. 8 4 x 29–48 s Factor the expression. 7. 4 x 2 x 2 5 x2 2x 1 29. 2x 12x 3 30. 5ab 8abc 31. x 2 7x 6 32. x 2 x 6 8. 5 3t 4 t2 2 2t t 3 33. x 2 2x 8 34. 2x 2 7x 4 9. 4x 1 3x 7 10. x x 1 x 2 2 2 35. 9x 2 36 36. 8x 2 10x 311. 2x 1 12. 2 3x 37. 6x 2 5x 6 38. x 2 10x 2513. y 4 6 y 5 y 2 39. t 3 1 40. 4t 2 9s 214. t 5 2t 3 8t 1 2 41. 4t 2 12t 9 42. x 3 2715. 1 2x x 2 3x 1 16. 1 x x2 43. x 3 2x 2 x 44. x 3 4x 2 5x 2s s s s s s s s s s s s s s s s s s s s 45. x 3 3x 2 x 3 46. x 3 2x 2 23x 6017–28 s Perform the indicated operations and simplify. 47. x 3 5x 2 2x 24 48. x 3 3x 2 4x 12 2 8x 9b 617. 18. s s s s s s s s s s s s s s s s s s s s 2 3b 49–54 s Simplify the expression. 1 2 1 119. 20. x2 x 2 2x 2 3x 2 x 5 x 3 x 1 x 1 49. 50. x2 3x 2 x 2 4 u 2 3 421. u 1 22. x2 1 x 3 5x 2 6x u 1 a2 ab b2 51. 52. 2 x 9x 8 x 2 x 12 x y x23. 24. 1 1 z y z 53. x 3 x2 9 2r s2 a b25. 26. s 6t bc ac
    • REVIEW OF ALGEBRA x 11 x 2 x 9 2x 4 an a 2n 154. 85. 86. x2 x 2 x2 5x 4 x3 a n 2s s s s s s s s s s s s s s s s s s s s 3 4 1 1 a b x y 87. 88.55–60 s Complete the square. a 5b 5 x y 155. x 2 2x 5 56. x 2 16x 80 89. 3 1 2 90. 961 557. x 2 5x 10 58. x 2 3x 1 91. 125 2 3 92. 64 4 359. 4x 2 4x 2 60. 3x 2 24x 50 93. 2x 2 y 4 3 2 94. x 5 y 3z 10 3 5 3s s s s s s s s s s s s s s s s s s s s 95. sy 6 5 96. (sa ) 461–68 s Solve the equation. 1 sx 5 8 97. 98.61. x 2 9x 10 0 62. x 2 2x 8 0 (st ) 5 4 sx 363. x 2 9x 1 0 64. x 2 2x 7 0 4 t 1 2sst 99. 100. sr 2n 4 1 4 sr 165. 3x 2 5x 1 0 66. 2x 2 7x 2 0 s2 3 s s s s s s s s s s s s s s s s s s s s67. x 3 2x 1 0 68. x 3 3x 2 x 1 0 101–108 s Rationalize the expression.s s s s s s s s s s s s s s s s s s s s sx 3 (1 sx ) 169–72 s Which of the quadratics are irreducible? 101. 102. x 9 x 169. 2x 2 3x 4 70. 2x 2 9x 4 x sx 8 s2 h s2 h71. 3x 2 x 6 72. x 2 3x 6 103. 104. x 4 hs s s s s s s s s s s s s s s s s s s s 2 1 105. 106.73–76 s Use the Binomial Theorem to expand the expression. 3 s5 sx sy 6 773. a b 74. a b 107. sx 2 3x 4 x 108. sx 2 x sx 2 x 2 4 2 575. x 1 76. 3 x s s s s s s s s s s s s s s s s s s s ss s s s s s s s s s s s s s s s s s s s 109–116 s State whether or not the equation is true for all77–82 s Simplify the radicals. values of the variable. 3 s 2 4 s32x 4 109. sx 2 x 110. sx 2 4 x 277. s32 s2 78. 3 79. 4 s54 s2 16 a a 1 111. 1 112. 1 1 x y s96a6 5 16 16 x y80. sxy sx y 3 81. s16a b 4 3 82. 5 s3a x 1 2 1 2 113. 114.s s s s s s s s s s s s s s s s s s s s x y 1 y 4 x 2 x83–100 Use the Laws of Exponents to rewrite and simplify s 115. x3 4 x7the expression. 116. 6 4x a 6 4x 4a83. 310 98 84. 216 410 16 6 s s s s s s s s s s s s s s s s s s s s
    • 12 s ANSWERS Answers q q q q q q q q q q q q q q q q q q q q q q q q q q1. 3a 2bc 2. 2x 3 y 5 3. 2x 2 10x 4. 4x 3x 2 9 s85 5 s13 63. 64. 1 2s2 65.5. 8 6a 6. 4 x 7. x 2 6x 3 2 68. 3t 2 21t 22 9. 12x 2 25x 7 7 s33 1 s510. x 3 x2 2x 11. 4x 2 4x 1 66. 67. 1, 68. 1, 1 s2 4 2 2 4 5 612. 9x 12x 4 13. 30y y y 69. Irreducible 70. Not irreducible14. 15t 2 56t 31 15. 2x 3 5x 2 x 1 71. Not irreducible (two real roots) 72. Irreducible16. x 4 2x 3 x2 2x 1 17. 1 4x 18. 3 2 b 73. a 6 6a 5b 15a 4b 2 20a 3b 3 15a 2b 4 6ab 5 b6 3x 7 2x u2 3u 119. 20. 21. 74. a 7 7a 6b 21a 5b 2 35a 4b 3 35a 3b 4 x2 2x 15 x2 1 u 1 2b 2 3ab 4a 2 x zx rs 21a 2b 5 7ab 6 b722. 23. 24. 25. a 2b 2 yz y 3t 75. x 8 4x 6 6x 4 4x 2 1 2 2 4 a c 3 2x 76. 243 405x 270x 90x 6 15x 8 x 1026. 27. 28. 29. 2x 1 6x 2 b2 c 2 2 x 1 77. 8 78. 3 79. 2 x 80. x 2 y30. ab 5 8c 31. x 6 x 1 32. x 3 x 2 81. 4a 2bsb 82. 2a 83. 3 26 84. 2 60 85. 16x 1033. x 4 x 2 34. 2x 1 x 4 a2 x y 2 135. 9 x 2 x 2 36. 4x 3 2x 1 86. a 2n 3 87. 88. 89. b xy s3 237. 3x 2 2x 3 38. x 5 1 90. 2 5s3 91. 25 92. 256 93. 2s2 x 3 y 6 239. t 1 t t 1 40. 2t 3s 2t 3s x3 141. 2t 3 2 42. x 3 x 2 3x 9 94. 9 5 6 95. y 6 5 96. a 3 4 97. t 5 2 98. y z x1 8 2 243. x x 1 44. x 1 x 2 t1 4 1 1 99. 100. r n 2 101. 102.45. x 1 x 1 x 3 46. x 3 x 5 x 4 s 1 24 sx 3 sx x47. x 2 x 3 x 4 48. x 2 x 3 x 2 x2 4x 16 2 103. 104. x 2 2x 1 x 1 x x 2 xsx 8 s2 h s2 h49. 50. 51. 52. x 2 x 2 x 8 x 4 3 s5 sx sy x 2 x 2 6x 4 105. 106.53. 54. 2 x y x2 9 x 1 x 2 x 4 3x 4 2x 5 2 107. 108. 57. ( x ) 2 2 1555. x 1 4 56. x 8 16 2 4 sx 2 3x 4 x sx 2 x sx 2 x 3 258. ( x ) 5 2 4 59. 2x 1 2 3 109. False 110. False 111. True 112. False60. 3 x 4 2 2 61. 1, 10 62. 2, 4 113. False 114. False 115. False 116. True