Book Extras: Principles of Algebra 2

eCourse

Don’t miss these video supplements that walk students through the course in an engaging, visual/auditory way!

Helpful Resources

  • Free Online Graphing Calculators – These calculators can be used instead of a graphing calculator for this course. (It is recommended that college-bound students get a graphing calculator, however, so they are already familiar with using one.)
    • Desmos (suggested, due to ease of identifying values and zooming in and out) – Note that you have to press the keyboard symbol in the bottom left to get a keyboard. Press y = and the values y equals, using x to stand for the independent variable. Use ^ to show exponents (use your right arrow to get back from an exponent to a regular value). Click on the graph of the curve itself to see the value at any particular point. Zoom in or out using the plus or minus signs. Add a second equation by clicking underneath where the first is typed and typing another one.
    • Math Is Fun – Use ^ to show exponents; use the first line for inputting the first equation, and the second for the second.
  • Free Online Matrix Calculators – These calculators can be used to calculate matrices.
    • Desmos – Input a new matrix for each matrix. Press A, A^-1, B to find (A^-1)(B).
    • Math Is Fun – Note that in order to calculate (A^-1)(B), you have to first press “inv(A)”, and then press the “to A” button in order to store it as Matrix A. Then press the AB button to calculate A times B, which is really (A^-1)(B), as you stored A^-1 as A (A^-1 is another way of saying the inverse of A).
  • The Atheist Delusion – Watch free online or order a DVD.

Corrections

Textbook

Page 152
The top equation on the page should be a(a – b) + b(a – b) = a^2 – ab + ba – b^2.

Page 359, Lesson 12.3
The first formula should have 1.005 in the dark blue, and the second 1 + 0.005. On the bottom line of each column, the parentheses should have 1.005 not 1.05.

With the formulas, the key is really just understanding what is meant by the letters. I would suggest just memorizing the P(subscript)i = (1+r)^t one and using it, only understand that the r has to be the rate used each time, and the t the number of times it’s compounded. Sometimes you’ll have to divide the rate given in a problem to find the r to use…or work to find the t to use.
For example, on Worksheet 12.3b, problem 2b, all the multiplying in the exponent was essentially just figuring out how many times in 4 years the compounding actually is done. So in this case, the rate used each time was 3% (the r), so added to 1 gave us 1.03 in the parentheses. And then in 4 years, it’d be compounded 8 times if it’s compounded every 6 months (so twice a year). So the t would be 8.
For another example, if the rate given is an annual rate and the interest is compounded monthly, first divide that annual rate by 12 to find the r for the formula. Likewise, if told to find the total after 10 years, multiply 10 by 12 to find the number of times that interest is compounded in 12 years (there are 12 months in a year).

Teacher Guide/Solution Manual

Worksheet 1.5, Problem 2
Answer should be approximately 4.376 yds.

Quiz 1, Problem 3d
The question should read “if the momentum is 12?” instead of “if the momentum is 10?” which would leave the solution correct except for all the 10s would be 12s in it. The answer would then be 8, as it is listed as right now. As it is right now, however the answer would be approximately 6.667, found by multiplying 2 by approximately 3.33.

Worksheet 1.6, Problem 3a
The 0.9144 in the solution should be 0.914. The answer should be 3.2822757111.

Worksheet 1.6, Problem 3b
3.28227571116 should be used instead of 3.28083989501; the answer should be 0.00186492938134.

Worksheet 1.6, Problem 3c
The answer should be approximately 6.714; in future printings, the instructions should be updated to say “rounding your answer to the second decimal place this time.” Given the updated instructions, the answer would then be 6.71, which would preserve significant digits (something not gone over in this course, but students who follow it should have their answers counted as correct).

Worksheet 1.7, Problem 3a
The answer should be approximately 295.405 yd/min.

Worksheet 2.6A, 1f and 2f
There should not be a +- sign in front of the answer.

Worksheet 2.7A, Problem 1i
The hint should be removed. This book follows the convention where all fractional exponents are defined as meaning the positive root, as that is how most books/calculators treat them. This is explained in the lesson book on page 69. So unless there’s a +- sign, the positive root is meant in this course.

Worksheet 2.7B, Problem 3a-3c
The answer to 3a is approximately 50.797 and the answer is 4 to 3b and 3c.

 Worksheet 2.7B, Problem 3
The instructions should say, “find the answer if the unknown equals 2” rather than “if a equals 2.”

Worksheet 3.2A, Problem 1b
Please ignore problem, as complex numbers haven’t been introduced yet.

Worksheet 3.2B, Problem 3a
The picture below shows how the simplification in the solution occurred in more depth.
Algebra

Worksheet 4.1, Problem 2d
The instructions should say “force” not “fore.”

Worksheet 4.2, Problem 2b
The problem is confusing as it is. Count as correct if students do not have a negative sign in their answer, as the scenario is unclear, or reword to “50 mi/hr in the negative direction and need to travel 300 mi in the negative direction, how long will it take you?” and remove the hint. Then the answer should be 6 hr.

Answer to Worksheet 4.7, Problem 4c
The answer should be equals, not approximately equals.

Worksheet 6.4, Problem 1
“A certain sum of money (P) at simple interest grows…”
should be
“A certain sum of money (P_i) at simple interest grows…”

Worksheet 11.4A, Problem 2b
The denominator in the answer should be -9.8, not positive 9.8.

Worksheet 11.4A, Problems 5c and 5d
Problems should reference 1e instead of 1d.

Worksheet 12.3b and Other Interest Problems
See textbook note about interest problems up above under textbook, page 359.

Worksheet 13.2, Problem 6b
Solution manual currently solves 5c instead of 5a. If solving 5a, the answer would be approximately 0.161/day; the second line of the solution would be r = ln(100/20)/10 yr = approximately 0.161/day.

Quiz 2 Grading
Problems 2b and 2c should only count as 5 points each.

Quiz 2, Problem 1f
Answer as the problem is should be a^4. Problem should be updated in future printings to be (a^2)^-2, in which case the current solution and answer is correct.

Quiz 2, Problem 1k
Problem should be clarified to be -(2)^2(-3)^2. Solution should be clarified like this: -(2)^2(-3)^2 = (-1)(2^2)(-3)^2 = (-1)(4)(9) = -36.

Quiz 7, Problem 1e
Answer should be r = 0 (all bold).

Quiz 7, Problem 1g
Answer should be (0,0) and the l = 0 should be removed. This note in the sidebar should be added: Found by graphing F(l) = 5l, which gets inputted into the calculator as y = 5x, and looking at the point where the line intercepts the x-axis.

Quiz 8, Grading
Each problem in sections 1 and 3 should be worth 3 instead of 4 points; those in section 5 should be worth 4 instead of 5 points.

Quiz 9, Grading
Each problem in Section 3 should be worth 4 instead of 5 points.

Quiz 10, Problem 2b
As it is right now, the answer of approximately +- sqrt 4/5i should not be listed, as it is complex and the instructions say not to list complex roots. However, students who list it should not be penalized. In future printings, the instructions should be adjusted to include complex roots, and the references to real solutions removed in the solution. If we were just looking for real roots, we could just graph to see the x = 0.

Test 2 (Clarification)
Note that students are not supposed to use a calculator that can solve algebraic expressions. This also means that section 2 should be solved with a different method besides matrices.
Test 3 (Clarification)
Make sections 1 ,3, 4, and 5 worth 3 points, each problem in section 2 worth 2 points (except the bonus question), and each problem on sections 6-10 worth 5 points.
Test 3, Problem 3b

Below is a more in-depth solution.

We start by factoring out a 10:
10(4(2)^3t + 3(2)^2t)
Then we factor out a 2^2t from each term inside the outer () above:
10(2)^2t(4(2^t) + 3)
Then we simplify, recognizing that 2^2 is 4 so 2^2t can simplify to 4^t:
10(4)^t(4(2^t) + 3)
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