A Math Learning Center publication adapted and arranged by EUGENE MAIER and LARRY LINNEN ALGEBRA THROUGH VISUAL PATTERNS, VOLUME A Math Learning Center Resource Copyright © 2005, 2004 by The Math Learning Center, PO Box 12929, Salem, Oregon 97309 Tel 503 370–8130 All rights reserved QP388 P0405 The Math Learning Center is a nonprofiit organization serving the education community Our mission is to inspire and enable individuals to discover and develop their mathematical confidence and ability We offer innovative and standards-based professional development, curriculum, materials, and resources to support learning and teaching To find out more visit us at www.mathlearningcenter.org The Math Learning Center grants permission to classroom teachers to reproduce blackline masters in appropriate quantities for their classroom use This project was supported, in part, by the National Science Foundation Opinions expressed are those of the authors and not necessarily those of the Foundation Prepared for publication on Macintosh Desktop Publishing system Printed in the United States of America ISBN 1-886131-60-0 ALGEBRA THROUGH VISUAL PATTERNS VOLUME Introduction vii LESSON Tile Patterns & Graphing LESSON Positive & Negative Integers 31 LESSON Integer Addition & Subtraction 47 LESSON Integer Multiplication & Division 57 LESSON Counting Piece Patterns & Graphs 73 LESSON Modeling Algebraic Expressions 91 LESSON Seeing & Solving Equations 113 LESSON Extended Counting Piece Patterns 135 VOLUME LESSON Squares & Square Roots 163 LESSON 10 Linear & Quadratic Equations 185 LESSON 11 Complete Sequences 217 LESSON 12 Sketching Solutions 251 LESSON 13 Analyzing Graphs 281 LESSON 14 Complex Numbers 315 Appendix 333 SQUARES & SQUARE ROOTS LESSON THE BIG IDEA Square roots are viewed as the lengths of sides of squares Methods of constructing a square of any given integral area, and thus the square root of any positive integer, are developed One of these constructions leads to the Pythagorean Theorem START-UP FOCUS FOLLOW-UP Overview Overview Overview Students construct squares of integral areas and establish the relationship between squares and square roots Students dissect squares and reassemble the pieces to form two squares and, conversely, dissect two squares and reassemble the pieces to form a single square In the process, they arrive at the Pythagorean Theorem They dissect rectangles and reassemble the pieces to form squares and, in so doing, construct square roots Students examine the relationship between products (quotients, sums) of square roots and square roots of products (quotients, sums) Students solve problems involving squares and square roots, using the Pythagorean Theorem as necessary They relate the arithmetic mean and the geometric mean of two positive numbers to the construction of squares Materials Centimeter grid paper (see Appendix), 2-3 sheets per student, transparency Start-Up Master 9.1, transparency Materials Follow-Up 9, copy per student Materials Centimeter grid paper (see Appendix), 2-3 sheets per student Scissors, pair per student Start-Up Master 9.1, copy per student and transparency Focus Masters 9.1-9.2, copy of each per student Focus Master 9.3, copy per student and transparency ALGEBRA THROUGH VISUAL PATTERNS | 163 TEACHER NOTES 164 | ALGEBRA THROUGH VISUAL PATTERNS SQUARES AND SQUARE ROOTS LESSON START-UP Overview Students construct squares of integral areas and establish the relationship between squares and square roots Materials Centimeter grid paper (see Appendix), to sheets per student, transparency Start-Up Master 9.1, transparency COMMENTS ACTIONS Distribute centimeter grid paper If your students are familiar with the basic properties of square roots, you may wish to omit this lesson to the students Tell them that square represents unit of area For each of the integers through 25 ask them to construct, if possible, a square whose vertices are grid intersection points and whose area is the given integer For each square they draw, ask the students to indicate its area and the length of its side Discuss SQUARES AND SQUARE ROOTS A student may believe they are finished when they have constructed all the squares whose sides lie along a gridline If this happens, you can simply tell the student there are more Normally, someone in the class will discover a square that “tilts.” Of the integers through 25, there are 13 for which a square exists that satisfies the conditions of Action Start-Up Master 9.1 attached at the end of this activity shows a square of each area A square of area 25 can also be obtained by carrying out a 3,4 pattern as described below LESSON START-UP BLACKLINE MASTER 9.1 One way to obtain a square that fits the conditions is to pick two intersection points as successive vertices In the instance shown below, one can get from point P to point Q by going units in one direction and in the other Repeating this 3,1 pattern, as shown, results in a square 5 10 10 3 1 13 17 17 13 16 Q 18 20 18 20 25 P Square generated by a 3,1 pattern continued next page ALGEBRA THROUGH VISUAL PATTERNS | 165 SQUARES AND SQUARE ROOTS LESSON START-UP COMMENTS ACTIONS continued A B D C Discuss with the students why they can be certain that the 13 integers mentioned in Comment are the only ones in the range through 25 for which squares exist that satisfy the conditions of Action The area of this square can be found by subtracting the area of the shaded regions from the area of the circumscribed square (see the figure) Note that regions A and C combine to form a rectangle of area as rectangles B and D Thus, the area of the inscribed square is 16 – 6, or 10 Since the area of the square is 10, the length of its side is 10 An approximation of 10 can be obtained by measuring the side of the square with a centimeter ruler If n is non-negative, the positive square root of n, written length of the side of a square of area n One way to see there are only 13 different areas is to note that if a square is to have area no greater than 25, then the distance between successive vertices must be less than or equal to Thus, if P and Q are successive vertices, Q must lie on or within a circle of radius whose center is at P In the sketch, the 13 intersections marked with an x are possibilities for Q that lead to 13 differently sized squares Any other choice for Q leads to a square the same size as one of these 13 x x x x x x x x x x x x x P 166 | ALGEBRA THROUGH VISUAL PATTERNS n, is the COMPLEX NUMBERS LESSON 14 FOCUS ACTIONS COMMENTS continued Note that no new colors—only scissors—are necessary to obtain square roots for i, that is, no new colors are needed to solve the equation x = i This is an illustration of the Fundamental Theorem of Algebra: Every polynomial equation with complex numbers as coefficients has solutions which are complex numbers 2 The square shown here has area Its edges are diagonals of unit squares 328 | ALGEBRA THROUGH VISUAL PATTERNS COMPLEX NUMBERS LESSON 14 FOLLOW-UP BLACKLINE MASTER 14 Compute: a) (3 – 5i) – (4 – 2i) b) (2 – 3i)(4 + i) + (1 – 2i)(5 – 3i) c) (22 – i) ÷ (2 – 3i) Find all solutions of the following equations: a) (x + 1) = –36 b) x + 10x + 100 = c) x – 3x + = The complex numbers can be associated with points in a coordinate plane by letting a + bi correspond to the point with coordinates (a, b) For example + 5i corresponds to the point (3, 5) Suppose a parallelogram has vertices at the origin O and at the points P and Q which correspond, respectively to – 3i and + 4i If OP and OQ are two sides of the parallelogram, find the coordinates of the fourth vertex, S, and the complex number corresponding to it How is the complex number corresponding to S related to the complex numbers corresponding to points P and Q? © THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 329 COMPLEX NUMBERS LESSON 14 ANSWERS TO FOLLOW-UP 14 a) A collection for – 5i consists of black and yellow tile; the opposite of a collection for – 2i consists of red and green tile Combining these collections and eliminating pairs of opposite tile yields a collection of red and yellow tile Hence, (3 – 5i) – (4 – 2i) = –1 –3i b) (2 – 3i)(4 + i) = 11 – 10i; (1 – 2i) (5 – 3i) = –1 – 13i (2 – 3i)(4 + i) + (1 – 2i) (5 – 3i) = 10 – 23i c) (22 – 7i) ÷ (2 – 3i) = + 4i b) c) a) x = –1 ± 6i 5x 25 x x2 5x x Completing the square: x + 10x + 100 = x + 10x + 25 = –75 (x + 5) = –75 = 25(–3) x + = ±5 i x = –5 ± i –3 –3x –3x x x2 x2 –3x x x2 x2 –3x x x –3 330 | ALGEBRA THROUGH VISUAL PATTERNS Multiplying by and then completing the square: x – 3x + = 4x – 12x + 36 = (2x – 3) = –45 = 9(–5) 2x – = ±3 i 2x = ± i x = (3 ± i)⁄ © THE MATH LEARNING CENTER COMPLEX NUMBERS LESSON 14 ANSWERS TO FOLLOW-UP 14 (CONT.) Coordinates of S are (11, 1) which is associated with the complex number 11 + i This is the sum of the numbers corresponding to P and Q Q (6,4) S (11,1) –1 10 11 12 –2 –3 P (5,–3) –4 © THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 331 TEACHER NOTES 332 | ALGEBRA THROUGH VISUAL PATTERNS APPENDIX ⁄ -Inch Grid © THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 333 APPENDIX 1-Centimeter Grid 334 | ALGEBRA THROUGH VISUAL PATTERNS © THE MATH LEARNING CENTER APPENDIX Coordinate Grids © THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 335 APPENDIX Red and Black Counting Pieces (print back-to-back with page 337) 336 | ALGEBRA THROUGH VISUAL PATTERNS © THE MATH LEARNING CENTER APPENDIX Back for Red and Black Counting Pieces © THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 337 APPENDIX Blank Counting Pieces 338 | ALGEBRA THROUGH VISUAL PATTERNS © THE MATH LEARNING CENTER APPENDIX Algebra Pieces (print back-to-back with page 340) APPENDIX Back for Algebra Pieces APPENDIX n-Frames (print back-to-back with page 342) APPENDIX Back for n-Frames 342 | ALGEBRA THROUGH VISUAL PATTERNS © THE MATH LEARNING CENTER ... Value, v (n ) : ? ?2 –1 n (? ?2) + (–1) (–1) + 0+1 1 +2 2+3 n + (n + 1) Arrangement No.: Value, v (n ) : ? ?2 –1 n 2( ? ?2) + 2( –1) + 2( 0) + 2( 1) + 2( 2) + 2( n ) + ALGEBRA THROUGH VISUAL PATTERNS |... 2n + 38 and 4n – 12 have the same value if 2n – 12 is 38 This is the case if n is 25 , that is, if n = or n = –5 38 n2 n2 38 n2 n2 n2 n2 – 12 continued next page ALGEBRA THROUGH VISUAL PATTERNS |... are 5 by 10 5 Perimeter = 20 + + 29 ≈ 17.5 cm 13 + 2 a = 2 + = 20 ; a = 20 b = 2 + = 13; b = 13 c=4 d = + 2 = 29 ; d = 29 4 14 ft h h2 62 14 + = h = 14 – = 160 h = 160 ≈ 12. 6 ft ft Area = x – – –