Announcements
Office hours this morning are delayed till approximately 11:05am this morning because I have to do a course observation in our department. This is the last one of the semester so after this office hours should be normal again. Alaso sorry for the repeated announcements on Brightspace and here, I just wanted to make sure everyone got the message.
I wanted to give some comments on the exam problems, besides their solutions (version A and version B). On #1, performance was generally strong on (a) and decent on (b). Most people on (b) realized that they needed to solve two different equations (\(3(x-4)=8\) and \(3(-(x-4))=8\)) but lots of people got it wrong what the two equations were. One mistake was to mess up the parentheses and solve \( 3x-4=8 \) instead of \(3(x-4)=8 \). Another was to solve \( 3(x-4)=8 \) and \( 3(x+4)=8 \). The best way to think about it is that \( \lvert x-4\rvert \) is either going to be equal to \( (x-4) \) or to \(-(x-4)\), so you need to replace \( \lvert x-4 \rvert \) by these two expressions and solve both resulting equations. I was also generally happy with the way people wrote their solutions, as a list of equations, one after another.
While #1 was an exercise of solving equations, problems #2 and #3 were exercises in transforming equations. This was from earlier in the class, and probably this is why people were less likely to write their solutions in the right way, as a chain of equalities. I say this all the time, but let me say it again: When you're solving equations, you write down a list of different equations, each derived from the last. The different terms in all these equations are not all equal to each other, so you don't get a single chain of equations. When you're transforming expressions into different forms, each transformed expression is equal to the last one. So, you get a single chain of equalities. You can write this chain of equalities all on one line, if it fits, or you can break the line at some point and start the next line with an equal sign. (Typically, people make the equal signs line up, because it looks nice and is easy to read.) One thing I saw people do a few times was to write arrows from one expression to the next. I appreciate that you were trying to indicate that the expression had been transformed into another one, but an arrow is a mathematical symbol already (it's used in a few different ways, none of which comes up in this class) and it does not mean this. The right symbol is an equal sign. (If you wanted to indicate that there's something different about the situation than when there is an equation to be solved, the right way to do it would be to preface the equation with, "For all x". That's what makes the difference: when you're solving an equation, the equation is only true for specific values of x and you're trying to figure out which. When you're transforming expressions, the equation holds for all choices of x.)
In 2(b) and 2(c), one step of the problem was to square an expression, computing \( (x^3+x^{-2})^2 \) in (b) and \( (\sqrt{x+2}+4)^2 \) in (c). It was very common for people to do this by just squaring each term separately, but this is wrong because it is not true that \( (a+b)^2=a^2 + b^2 \). I have to be harsh in my grading here, because nobody should pass an algebra class and make this mistake. The correct version is \( (a+b)^2=a^2+2ab+b^2\), which in 2(b) gives you \[ (x^3+x^{-2})^2 = (x^3)^2 + 2x^3x^{-2}+(x^{-2})^2 = x^6+2x + x^{-4} \] and in 2(c) gives you \[ \begin{aligned} \bigl(\sqrt{x+2} + 3\bigr)^2 &= (\sqrt{x+2})^2 + 2\sqrt{x+2}(3) + 3^2\\ &=x+2 + 6\sqrt{x+2} + 9 = x+11+6\sqrt{x+2}. \end{aligned} \]
In 2(b), using the work above, we have \[ (x^3+x^{-2})^2 + x = x^6+2x + x^{-4} + x = x^6+3x+x^{-4}.\] The problem statement says to get rid of all negative exponents, which means writing \( x^{-4} \) as \( \frac{1}{x^4} \), giving a final answer of \[ x^6+3x+\frac{1}{x^4}. \] There was a very common mistake in this step, which was a bit mystifying to me, of instead putting \[ \frac{x^6 + 3x}{x^4}. \] In this answer, instead of adding \(x^{-4}\) to \(x^6+3x\), we've multiplied \(x^{-4}\) by \(x^6+3x\).
Problem #4 was basically the same problem as on the quiz. It reduces to the problem of finding height and width that multiply to 2000 (or 1800 in version B of the exam) and add to 100. The majority of people got to this point. Unlike on the quiz, I took care to make it so that you couldn't get the right answer just by going through the factors of 2000. Instead, you need to use the two equations \(xy=2000\) and \(2x+2y=200 \) to get a quadratic equation, which you have to solve using the quadratic formula.
People did well on problems #5 and #6. The only common mistake on #6 was messsing up the ordering with the negative slope, which is the smallest. (Negative numbers are smaller than positive ones.) I guess you understand lines well, which is good.
For #7, you need to know what the domain of a function is: the values that you can plug into it. Here, the only reason you wouldn't be able to evaluate the function is if the denominator is zero, which happens when \( x^2+3x =0 \) in version A of the problem. You have to solve this equation, which you can do by factoring to obtain \( x(x+3)=0 \), giving the solutions \( x =0\) and \(x=-3\). Then the domain consists of all real numbers except these two. You can just write that sentence. Or you could write \( (-\infty,-3)\cup(-3,0)\cup(0,\infty) \), which is perfectly valid if perhaps a bit less clear. (If you really want to write the answer purely in mathematical symbols, the best way would be to write \( \mathbb{R} \setminus \{-3,0\} \). The symbol \( \mathbb{R} \) means the real numbers, and the symbol \( \setminus\) is an operator along the lines of \(\cup\) or \(\cap\), but it means to remove the second set from the first. So \( \mathbb{R} \setminus \{-3,0\} \) means the set of real numbers with -3 and 0 removed.)
Problem #8 was probably the one that people did the worst on. This was illuminating to me, and I'll talk more in class about graphing and about piecewise functions. Problem #9 was not exactly easy, but it was the same basic problem we practiced a few times before the exam.
What does your exam score mean for you? Well, in the most prosaic sense, see the gradescale in the post below. It should give you an accurate sense of where you stand in the class. But let me try to give some more advice-like advice.
First, if you did well on the exam, congratulations! I'm very happy for you.
If you're not happy with your performance on the exam, the first thing to do is to get your exam back and look at where you went wrong. It is very painful to look back at all the red ink—I remember getting back exams I'd messed up in college and refusing to look at them—but please do it anyway. It is important to see where you made mistakes and read any comments that I wrote to you.
Next, if you did really badly on this exam—I'm thinking of scores in the teens and below, out of 61 points—you should probably drop the class. It's not that I want you to drop—I'd much prefer that you stay in the class and excel in the remaining exams—but if you can't do any of the problems now, it's probably too late for you. You will probably fail, and it will be a poor use of your time to work and go to class if you're going to fail. You're better off trying again in a different semester, perhaps when you have more time to work on it. I'm not saying that everyone in this category is definitely going to fail the class, but if you want to remain and do better, you should think honestly for yourself and have a good reason to believe that you'll do better going forward. Note that even if you plan to spend more time and put more effort into the class going forward, it's difficult to succeed in a math class once you've fallen behind, since it's hard to follow the new material as it comes if you don't know the old material.
If you're somewhere in the middle—doing worse than you'd like but not at the very bottom—you can improve. The way to do it is to put more time and effort into solving the homework questions and taking practice exams. Please note that you should not just put more effort into, say, reading the textbook or watching videos. Math is something that you have to do. Reading the textbook is great, if you're doing it in the service of learning how to solve problems that you were stuck on. But if you just sit down and read the textbook without doing any problems, you won't learn anything.
One very good resource is our department's Tutoring Center. Try to do homework problems, and go to Tutoring Center and get help on the ones that you're stuck on. If you're diligent about this, you can really learn a lot of material this way in a short time.
I've posted your midterm scores on Brightspace. You can interpret your score as follows:
- 45–61: A- to A
- 35–44: B- to B+
- 27–34: C to C+
Anything lower than that is a D or an F.
In the interest of transparency, and also because I hate answering questions about I assign grades, let me explain in as much detail as possible how I assign grades for the class. For each midterm and final, after I grade the test, I make a gradescale as above. I don't set the scale to make a certain number of A's, B's, and C's. Rather, I look at the tests and try to judge what scores match my expectations for what merits an A, B, or C. This means that if everyone in the class does well, I set the gradescale so that there are more A's.
When it comes time to assign final grades for the course, I use the gradescales to convert the midterm and final scores to a numerical scale of 0–100. In more detail, I convert the low end of the A range to 90, the low end of the B range to 80, and the low end of the C range to 70. I set the high end of the A range to something above 100. For scores in the middle of a range, I interpolate linearly to convert them to the numerical scale. As a consequence, even though the gradescale for this exam says that 44 is a B+ and 45 is an A-, there is no big difference in getting a 44 rather than a 45 on this exam. When converted to the numerical scale, you'd just be getting an 89 instead of a 90 or something like that.
For the homework and quiz part of the grade, I also convert your scores to the numerical grade, though I'm a bit more ad hoc about it. Finally, I take all your numerical grades and compute a weighted average according to the weights given in the syllabus and arrive at your final numerical grades. Then, I make the final letter grades based on these. Roughly speaking, I'd convert numerical grades of 90 and above to an A or A-, grades of 80 to 90 to a B+, B, or B-, etc., but I use my judgment when setting the cutoffs (it's not like I'm giving people who got a numerical grade of 89.9 a B+).
One very special grading policy that I have: If you ask me whether I grade on a curve, I will ask you questions about what you mean by grading on a curve, probably until you give up asking your question. (If you can explain to me what you mean by grading on a curve, I am happy to answer the question, but it is not very common that anyone has a clear sense of what it means to grade on a curve.) I do hope that my explanation here is enough that you'll understand how the grading works.
I think people have already asked me if there's any extra credit available to raise your grade. The short answer is no (see the syllabus). It's not that I want to be harsh. In fact, I really want you all to get good grades. The reason that I don't give any extra credit opportunities is that I think that the exams are a fair assessment of what you know and how you're able to perform. If you're unhappy with how you did on this exam, the solution is to do better on the remaining exams, which are weighted more heavily because your performance at the end of the course is more important than your performance at the beginning. In fact, if you do show significant improvement from the beginning to the end of the class, sufficiently so that your earlier exams don't accurately reflect your ability at the end of the course, I increase your grade by a notch (see the syllabus, and note that this is not a common occurrence).
Here's one last note about how to write your algebra solutions. Earlier in the class, I divided up algebra into two parts: transforming expressions, and solving equations. The last post here was about solving equations. I wrote that when solving an equation like \( 3x+1=5 \), your answer should look like a list of equations, each derived from the last, like this:
Today, I want to talk about transforming expressions. I've talked before about how to do this, and it was basically the topic of the first quiz, but I think I should repeat myself now about it. Transforming an expression is when you take something like \( (s+1)(s^2-2s-5) \) and expand and simplify it. When you do this, you should be writing a chain of equalities, like this:
\[ \begin{aligned} (s+1)(s^2-2s-5) &= s(s^2-2s-5) + 1(s^2-2s-5)\\ &= s^3 - 2s^2 - 5s + s^2 - 2s - 5\\ &= s^3 + (-2s^2 + s^2) + (-5s-2s) - 5\\ &= s^3 + (-2+1)s^2 + (-5-2)s - 5\\ &= s^3 - s^2 - 7s - 5 \end{aligned} \]What I've written here is six expressions all connected by equal signs. It's written on six lines, with the last five starting with an equal sign, but that's just for typographical reasons. It could equally well be written on one line. There's nothing special about starting the line with an equal sign. It's just a way to continue the chain of equations on a new line. Also, I should note that I've written out this solution in great detail, but that it's fine (and expected) that your solution would compress some of these steps together, perhaps like this:
\[ (s+1)(s^2-2s-5) = s^3 - 2s^2 - 5s + s^2 - 2s - 5 = s^3 - s^2 - 7s - 5. \]In this situation where we're starting with an expression and transforming it, we end up with a chain of expressions that are all equal to each other. Note that this is not the case when you're solving an equation. Looking back at the example of solving an equation, we start with \(3x + 1 =5 \) and then on the next line have \(3x = 4 \). It wouldn't make sense to tie these two equations together with an equal sign into a single equation \(3x+1=5=3x=4\), because while \(3x+1=5\) and \(3x=4\) are true, it is not the case that \(3x+1=3x\) or \(5=4\). Lots of people did write stuff like this on their quizzes, because of confusing the situation between transforming expressions and solving equations, and I hope this clarifies the difference for you.
To summarize: Transforming an expression? Write a chain of equalities, all connected by equal signs. Solving an equation? Write one equation after the next.
I've graded the quizzes and have a lot of comments on how I want you to do your algebra. I want to try to make myself very clear here. I'll focus on Problem 1 on the quiz, which was to find all solutions to \( \lvert 3x + 1 \rvert = 5 \). The basic idea for finding the solutions is that you need to solve the equations \( 3x + 1 = 5 \) and \( 3x + 1 = -5 \). Take a look at the solutions to see the answer in more detail.
Many people wrote answers that look like this:
It was clear from these answers that you solved these equations perfectly; they were also quite nice looking, aesthetically. Nonetheless, solutions like that violate the instructions at the top of the exam: When solving equations, write down one equation after another. Do not cross out or write on top of an equation. Specifically, you can see that the equation \( 3x = 4 \) was written, and then fractions \( \frac{}{3} \) were written on top of it. (The \( -1 \) and the line across indicating that you're writing the result of adding \( -1 \) to each side of the fraction isn't literally writing on top of the equation, but I also don't want you to do it.) I did take away points for writing your solution this way and will continue to do so. I want your solutions to look something like this:
Since most people in the class wrote solutions in this way, it must be that you were taught to write solutions this way. So I owe you a detailed explanation of why I'm making you do it differently, which I'll try to provide now in question/answer format.
Why don't you want me to write it that way? For lots of reasons. Practically speaking, it is a bad way to carry out math if you'd like to get the answer right. Let me elaborate:
- Let's say you're solving the equation \( \frac{x}{2} + 7 = 18 \).
I cannot tell you how many times I've seen solutions like:
We are supposed to multiply both sides by 2, but here the multiplication is read as only applying to the first term on the left-hand side. You can make a mistake however you're writing things, but I think it's easier to make a mistake when you write on top of equations or cross things out. The equations get messier and harder to read, and it's very easy to neglect parentheses. - Your solution should be a record leading from your starting equation to the changed, equivalent equations you get to along the way. If you write on top of them or cross things out, we lose the record.
- Philosophically, when we write math, we're writing down a logical argument. The story is that we're trying to find solutions to \( 3x + 1 = 5 \), and we know that \( 3x = 4 \) has the same solutions, and then we know that \( x = \frac43 \) has the same solutions, and now we're down to an equation whose solutions (well, there's only one) are incredibly obvious. Math papers really are written like this, with much more English text than equations. I don't expect you to be quite so verbose when you're giving answers to a quiz, but writing down those three equations one after another is a good, quick way of giving that story. Writing down the equations and then writing on top of them is not.
Okay, but don't you want me to show my work? In the original version, it's clearly indicated that we're subtracting 1 from both sides of the equation or dividing both sides of the equation by 3. In the version you're asking for, it doesn't say how we get from one equation to the next. True, it doesn't! But since all the steps are just multiplying both sides of the equation by the same quantity or adding the same thing to both sides of the equation, it's pretty obvious what you did. Also, note that your solution also obscures what you did. We see the equation \( \frac{3x}{3} = \frac43 \) but we don't see the intermediate equation \(3x = 4 \) anymore, since it got written on top of.
If you really want to explain each step of the solution, a good way to do it would be to annotate it with sentences in English, like:
Start with \[ 3x + 1 = 5.\] Subtract 1 from both sides: \[ 3x = 4. \] Divide both sides by 3: \[ x = \frac43.\]
Math papers really are written like this. It's kind of overkill for doing algebra in a class like this. I could also imagine some notation where you have a sort of table, and on the left side you write the equations and on the right you give the explanation of what you did to get to that step. That would also be quite clear, but also overkill for writing solutions on an exam.
I am annoyed that people have been telling me to do it one way for my entire life and now you're telling me to do it differently. I can't argue with that! But I think the issue is with high-school math education and not with what I'm telling you now. No math professor wants you to write algebra the way you've been told. No practitioner of mathematics writes algebra that way. Math professors grading exams are always complaining about students doing it!
Update: I fixed another typo in the solution to #2, where it said the answer was \( (-\infty,3)\cup\{5\} \) when it should have been \( (-\infty,3]\cup\{5\} \).
I updated the quiz solutions with a better solution to #2. The original solution was less clear. And also it was not correct, because it missed the solution \( x=5\)! I meant to ask the slightly easier question of finding all the solutions to \( (x-5)^2(x+3) < 0 \) rather than \( (x-5)^2(x+3) \leq 0 \). But I am not penalizing anyone for failing to find the solution \(x=5\).