Bob Miller's Algebra for the Clueless (Clueless Series)

Bob Miller's Algebra for the Clueless (Clueless Series)

Language: English

Pages: 288

ISBN: 0071473661

Format: PDF / Kindle (mobi) / ePub


A is for Algebra-and that's the grade you'll pull when you use Bob Miller's simple guide to the math course every college-bound kid must take

With eight books and more than 30 years of hard-core classroom experience, Bob Miller is the frustrated student's best friend. He breaks down the complexities of every problem into easy-to-understand pieces that any math-phobe can understand-and this fully updated second edition of Bob Miller's Algebra for the Clueless covers everything a you need to know to excel in Algebra I and II.

Principles of Linear Algebra with Mathematica (Pure and Applied Mathematics: A Wiley Series of Texts, Monographs and Tracts)

Functional Analysis: An Introduction (Graduate Studies in Mathematics)

A First Course in Graph Theory (Dover Books on Mathematics)

Real and Convex Analysis (Undergraduate Texts in Mathematics)

 

 

 

 

 

 

 

 

 

 

 

 

 

ᎏᎏ a2 b2 ᎏ1ᎏ × ᎏ a2b2 ᎏ + ᎏ1ᎏ × ᎏ a2b2 ᎏ = a 1 b 1 ᎏᎏᎏ = ab2 + a2b ᎏᎏ = ab(b + a) ᎏᎏ 1 b2 1 b2 b2 − a2 (a − b)(a + b) ᎏᎏ × ᎏ a2ᎏ − ᎏᎏ × ᎏa2ᎏ a2 1 b2 1 = ab ᎏ a − b Now, let’s relate the beginning of the chapter to the middle of the chapter. Fractional Exponents We would like to relate fractional exponents to radicals. R a d i c a l s a n d E x p o n e n t s 129 DEFINITION x1/r = ͙ r xෆ where x is not negative if r is even 81/3 = ͙ 3 8ෆ = 2 251/2 = ͙2ෆ5ෆ =

A M P L E 2 — Solve for x, using the quadratic formula: 4x2 + 12x + 9 = 0. a = 4 b = 12 c = 9 −b Ϯ ͙bෆ2ෆ−ෆ 4 ෆaෆcෆ −12 Ϯ ͙1ෆ2ෆ2ෆ−ෆ 4 ෆ(4 ෆ)(ෆ9ෆ)ෆ x = ᎏᎏ = ᎏᎏᎏ 2a 2(4) −12 Ϯ ͙1ෆ4ෆ4ෆ − ෆ 1 ෆ4ෆ4ෆ −12 Ϯ 0 = ᎏᎏᎏ = ᎏ = − 3 ᎏ 2(4) 8 2 The equation has two equal roots, both −3/2. E X A M P L E 3 — Solve for x, using the quadratic formula: x2 − x + 8 = 0. 150 B O B M I L L E R ’ S A L G E B R A F O R T H E C L U E L E S S a = 1 b = −1 c = 8 −b Ϯ ͙bෆ2ෆ−ෆ 4 ෆaෆcෆ −(−1) Ϯ

E X A M P L E 2 — Solve for all values of x: ͙4ෆxෆ + 3 = 0. ͙4ෆxෆ + 3 = 0 ͙4ෆxෆ = −3 NOTE 4x = 9 9 You must check. Some- x = ᎏ 4 times the answer(s) will To check: ͙4ෆ(9 ෆ/4 ෆ)ෆ + 3 ՘ 0. 3 + 3 ≠ 0. The equation has check, sometimes not. no solution. 154 B O B M I L L E R ’ S A L G E B R A F O R T H E C L U E L E S S E X A M P L E 3 — Solve for all values of x: ͙2ෆxෆ + ෆ 1 ෆ + x = 7. Isolate the square root. ͙2ෆxෆ + ෆ 1 ෆ + x = 7 Square both sides. ͙2ෆxෆ + ෆ 1 ෆ = 7 −

subtracting are done the same way. So we’ll concentrate on addition. a b a + b 3 2 Rule: ᎏ + ᎏ = ᎏ ᎏ + ᎏ = 5 ᎏ c c c 7 7 7 F r a c t i o n s , D e c i m a l s , P e r c e n t s , a n d G r a p h s 251 The problem is when the bottoms are different. We Multiple (nn): Take all nn must find the least common denominator, which in and multiply by a nn. Mul- reality is the least common multiple (LCM). tiples of 5: 5 × 1, 5 × 2, 5 × 3, 5 × 4, 5 × 5, 5 × 6, . . . = 5, 10, 15, 20,

in the set. equal.” DEFINITION Equal sets: Set A = Set B if they have exactly the same elements. E X A M P L E 5 — A = {5, 7} and B = {7, 5}. A = B. Order does not matter. E X A M P L E 6 — D = {1, 1, 1, 2, 2, 1, 2} and E = {1, 2}. D = E. Repeated elements don’t count. Set D has only two elements. DEFINITION A ʜ B is read: “A union B” and is the set of elements in A or in B or in both. DEFINITION A ʝ B is read: “A intersect (ion) B” and is the set of elements in both sets. S e t

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