2. Algebra Errors
The topics covered here are errors that students often make in doing algebra, and not just errors typically made in an algebra class. I’ve seen every one of these mistakes made by students in all level of classes, from algebra classes up to senior level math classes! In fact, a few of the examples in this section will actually come from calculus.
If you have not had calculus you can ignore these examples. In every case where I’ve given examples I’ve tried to include examples from an algebra class as well as the occasion example from upper level courses like Calculus.
I’m convinced that many of the mistakes given here are caused by people getting lazy or getting in a hurry and not paying attention to what they’re doing. By slowing down, paying attention to what you’re doing and paying attention to proper notation you can avoid the vast majority of these mistakes!
Division by Zero
Everyone knows that \(\frac{0}{2} = 0\) the problem is that far too many people also say that \(\frac{2}{0} = 0\) or \(\frac{2}{0} = 2\)! Remember that division by zero is undefined! You simply cannot divide by zero so don’t do it!
Here is a very good example of the kinds of havoc that can arise when you divide by zero. See if you can find the mistake that I made in the work below.
- \(a = b\)We’ll start assuming this to be true.
- \(ab = {a^2}\)Multiply both sides by a.
- \(ab - {b^2} = {a^2} - {b^2}\)Subtract \({b^2}\) from both sides.
- \(b\left( {a - b} \right) = \left( {a + b} \right)\left( {a - b} \right)\)Factor both sides.
- \(b = a + b\)Divide both sides by \(a - b\).
- \(b = 2b\)Recall we started off assuming \(a = b\).
- \(1 = 2\)Divide both sides by b.
So, we’ve managed to prove that 1 = 2! Now, we know that’s not true so clearly we made a mistake somewhere. Can you see where the mistake was made?
The mistake was in step 5. Recall that we started out with the assumption \(a = b\). However, if this is true then we have \(a - b = 0\)! So, in step 5 we are really dividing by zero!
That simple mistake led us to something that we knew wasn’t true, however, in most cases your answer will not obviously be wrong. It will not always be clear that you are dividing by zero, as was the case in this example. You need to be on the lookout for this kind of thing.
Remember that you CAN’T divide by zero!
Bad/lost/Assumed Parenthesis
This is probably error that I find to be the most frustrating. There are a couple of errors that people commonly make here.
The first error is that people get lazy and decide that parenthesis aren’t needed at certain steps or that they can remember that the parenthesis are supposed to be there. Of course, the problem here is that they often tend to forget about them in the very next step!
The other error is that students sometimes don’t understand just how important parentheses really are. This is often seen in errors made in exponentiation as my first couple of examples show.
\({\left( {4x} \right)^2} = {\left( 4 \right)^2}{\left( x \right)^2} = 16{x^2}\) \(4{x^2}\)
Note the very important difference between these two! When dealing with exponents remember that only the quantity immediately to the left of the exponent gets the exponent. So, in the incorrect case, the x is the quantity immediately to the left of the exponent so we are squaring only the x while the 4 isn’t squared. In the correct case the parenthesis is immediately to the left of the exponent so this signifies that everything inside the parenthesis should be squared!
Parenthesis are required in this case to make sure we square the whole thing, not just the x, so don’t forget them!
\({\left( { - 3} \right)^2} = \left( { - 3} \right)\left( { - 3} \right) = 9\) \( - {3^2} = - \left( 3 \right)\left( 3 \right) = - 9\)
This one is similar to the previous one, but has a subtlety that causes problems on occasion. Remember that only the quantity to the left of the exponent gets the exponent. So, in the incorrect case ONLY the 3 is to the left of the exponent and so ONLY the 3 gets squared!
Many people know that technically they are supposed to square -3, but they get lazy and don’t write the parenthesis down on the premise that they will remember them when the time comes to actually evaluate it. However, it’s amazing how many of these folks promptly forget about the parenthesis and write down -9 anyway!
\(\begin{eqnarray*}{x^2} + 3x - 5 - \left( {4x - 5} \right) & = & {x^2} + 3x - 5 - 4x + 5\\ & = & {x^2} - x\end{eqnarray*}\) \({x^2} + 3x - 5 - 4x - 5 = {x^2} - x - 10\)
Be careful and note the difference between the two! In the first case I put parenthesis around then \(4x - 5\) and in the second I didn’t. Since we are subtracting a polynomial we need to make sure we subtract the WHOLE polynomial! The only way to make sure we do that correctly is to put parenthesis around it.
Again, this is one of those errors that people do know that technically the parenthesis should be there, but they don’t put them in and promptly forget that they were there and do the subtraction incorrectly.
\(\sqrt {7x} = {\left( {7x} \right)^{\frac{1}{2}}}\) \(\sqrt {7x} = 7{x^{\frac{1}{2}}}\)
This comes back to same mistake in the first two. If only the quantity to the left of the exponent gets the exponent. So, the incorrect case is really \(7{x^{\frac{1}{2}}} = 7\sqrt x \) and this is clearly NOT the original root.
\(\begin{eqnarray*} -3\int{6x-2\,dx} & = & -3\left( 3{{x}^{2}}-2x \right)+c \\ & = &-9{{x}^{2}}+6x+c \end{eqnarray*}\) \(\begin{eqnarray*} -3\int{6x-2\,dx} & = & -3\cdot 3{{x}^{2}}-2x+c \\ & = & -9{{x}^{2}}-2x+c \end{eqnarray*}\)
Note the use of the parenthesis. The problem states that it is -3 times the WHOLE integral not just the first term of the integral (as is done in the incorrect example).
Improper Distribution
Be careful when using the distribution property! There two main errors that I run across on a regular basis.
\(4\left( {2{x^2} - 10} \right) = 8{x^2} - 40\) \(4\left( {2{x^2} - 10} \right) = 8{x^2} - 10\)
Make sure that you distribute the 4 all the way through the parenthesis! Too often people just multiply the first term by the 4 and ignore the second term. This is especially true when the second term is just a number. For some reason, if the second term contains variables students will remember to do the distribution correctly more often than not.
\(\begin{eqnarray*}3{\left( {2x - 5} \right)^2} & = & 3\left( {4{x^2} - 20x + 25} \right)\\ & = & 12{x^2} - 60x + 75\end{eqnarray*}\) \(\begin{eqnarray*}{3\left( {2x - 5} \right)^2} & = & {\left( {6x - 15} \right)^2}\\ & = & 36{x^2} - 180x + 225\end{eqnarray*}\)
Remember that exponentiation must be performed BEFORE you distribute any coefficients through the parenthesis!
Additive Assumptions
I didn’t know what else to call this, but it’s an error that many students make. Here’s the assumption. Since \(2\left( {x + y} \right) = 2x + 2y\) then everything works like this. However, here is a whole list in which this doesn’t work.
\[{\left( {x + y} \right)^2} \ne {x^2} + {y^2}\] \[\sqrt {x + y} \ne \sqrt x + \sqrt y \] \[\frac{1}{{x + y}} \ne \frac{1}{x} + \frac{1}{y}\] \[\cos \left( {x + y} \right) \ne \cos x + \cos y\]It’s not hard to convince yourself that any of these aren’t true. Just pick a couple of numbers and plug them in! For instance,
\[\begin{eqnarray*}{\left( {1 + 3} \right)^2} & \ne & {1^2} + {3^2}\\{\left( 4 \right)^2} & \ne & 1 + 9\\16 & \ne & 10\end{eqnarray*}\]You will find the occasional set of numbers for which one of these rules will work, but they don’t work for almost any randomly chosen pair of numbers.
Note that there are far more examples where this additive assumption doesn’t work than what I’ve listed here. I simply wrote down the ones that I see most often. Also, a couple of those that I listed could be made more general. For instance,
\[{\left( {x + y} \right)^n} \ne {x^n} + {y^n} \hspace{0.5in} \mbox{for any integer } n \ge 2\] \[\sqrt[n]{{x + y}} \ne \sqrt[n]{x} + \sqrt[n]{y} \hspace{0.5in} \mbox{for any integer }n \ge 2\]Canceling Errors
These errors fall into two categories. Simplifying rational expressions and solving equations. Let’s look at simplifying rational expressions first.
Notice that in order to cancel the \(x\) out of the denominator I first factored an \(x\) out of the numerator. You can only cancel something if it is multiplied by the WHOLE numerator and denominator, or if IS the whole numerator or denominator (as in the case of the denominator in our example).
Contrast this with the next example which contains a very common error that students make.
Far too many students try to simplify this as,
\[3{x^2} - x \hspace{0.75in} \mbox{OR} \hspace{0.75in} 3{x^3} - 1\]In other words, they cancel the \(x\) in the denominator against only one of the \(x\)’s in the numerator (i.e. cancel the \(x\) only from the first term or only from the second term). THIS CAN’T BE DONE!!!!! In order to do this canceling you MUST have an \(x\) in both terms.
To convince yourself that this kind of canceling isn’t true consider the following number example.
This can easily be done just be doing the arithmetic as follows
\[\frac{{8 - 3}}{2} = \frac{5}{2} = 2.5\]However, let’s do an incorrect cancel similar to the previous example. We’ll first cancel the two in the denominator into the eight in the numerator. This is NOT CORRECT, but it mirrors the canceling that was incorrectly done in the previous example. This gives,
\[\frac{{8 - 3}}{2} = 4 - 3 = 1\]Clearly these two aren’t the same! So you need to be careful with canceling!
Now, let’s take a quick look at canceling errors involved in solving equations.
Too many students get used to just canceling (i.e. simplifying) things to make their life easier. So, the biggest mistake in solving this kind of equation is to cancel an \(x\) from both sides to get,
\[2x = 1 \hspace{0.25in} \Rightarrow \hspace{0.25in} x = \frac{1}{2}\]While, \(x = \frac{1}{2}\) is a solution, there is another solution that we’ve missed. Can you see what it is? Take a look at the next example to see what it is.
Here’s the correct way to solve this equation. First get everything on one side then factor!
\[\begin{eqnarray*}2{x^2} - x & = & 0\\ x\left( {2x - 1} \right) & = & 0\end{eqnarray*}\]From this we can see that either,
\[x = 0 \hspace{0.5in} \mbox{OR} \hspace{0.5in} 2x - 1 = 0\]In the second case we get the \(x = \frac{1}{2}\) we got in the first attempt, but from the first case we also get \(x = 0\) that we didn’t get in the first attempt. Clearly \(x = 0\) will work in the equation and so is a solution!
We missed the \(x = 0\) in the first attempt because we tried to make our life easier by “simplifying” the equation before solving. While some simplification is a good and necessary thing, you should NEVER divide out a term as we did in the first attempt when solving. If you do this, you WILL lose solutions.
Proper Use of Square Root
There seems to be a very large misconception about the use of square roots out there. Students seem to be under the misconception that
\[\sqrt {16} = \pm \,4\]This is not correct however. Square roots are ALWAYS positive or zero! So the correct value is
\[\sqrt {16} = 4\]This is the ONLY value of the square root! If we want the -4 then we do the following
\[ - \sqrt {16} = - \left( {\sqrt {16} } \right) = - \left( 4 \right) = - 4\]Notice that I used parenthesis only to make the point on just how the minus sign was appearing! In general, the middle two steps are omitted. So, if we want the negative value we have to actually put in the minus sign!
I suppose that this misconception arises because they are also asked to solve things like \({x^2} = 16\). Clearly the answer to this is \(x = \pm \,4\) and often they will solve by “taking the square root” of both sides. There is a missing step however. Here is the proper solution technique for this problem.
\[\begin{eqnarray*}{x^2} & = & 16\\x & = & \pm \sqrt {16} \\x & = & \pm \,4\end{eqnarray*}\]Note that the \( \pm \) shows up in the second step before we actually find the value of the square root! It doesn’t show up as part of taking the square root.
I feel that I need to point out that many instructors (including myself on occasion) don’t help matters in that they will often omit the second step and by doing so seem to imply that the \( \pm \) is showing up because of the square root.
So, remember that square roots ALWAYS return a positive answer or zero. If you want a negative you’ll need to put it in a minus sign BEFORE you take the square root.
Ambiguous Fractions
This is more a notational issue than an algebra issue. I decided to put it here because too many students come out of algebra classes without understanding this point. There are really three kinds of “bad” notation that people often use with fractions that can lead to errors in work.
The first is using a “/” to denote a fraction, for instance 2/3. In this case there really isn’t a problem with using a “/”, but what about 2/3\(x\)? This can be either of the two following fractions.
\[\frac{2}{3}x \hspace{0.5in} \mbox{OR} \hspace{0.5in} \frac{2}{{3x}}\]It is not clear from 2/3\(x\) which of these two it should be! You, as the student, may know which one of the two that you intended it to be, but a grader won’t. Also, while you may know which of the two you intended it to be when you wrote it down, will you still know which of the two it is when you go back to look at the problem when you study?
You should only use a “/” for fractions when it will be clear and obvious to everyone, not just you, how the fraction should be interpreted.
The next notational problem I see fairly regularly is people writing,
\[\frac{2}{3}\,\,\begin{array}{*{20}{c}}{}\\ x\end{array}\]It is not clear from this if the \(x\) belongs in the denominator or the fraction or not. Students often write fractions like this and usually they mean that the \(x\) shouldn’t be in the denominator. The problem is on a quick glance it often looks like it should be in the denominator and the student just didn’t draw the fraction bar over far enough.
If you intend for the \(x\) to be in the denominator then write it as such that way, \(\frac{2}{{3x}}\), i.e. make sure that you draw the fraction bar over the WHOLE denominator. If you don’t intend for it to be in the denominator then don’t leave any doubt! Write it as \(\frac{2}{3}x\).
The final notational problem that I see comes back to using a “/” to denote a fraction, but is really a parenthesis problem. This involves fractions like
\[\frac{{a + b}}{{c + d}}\]Often students who use “/” to denote fractions will write this is fraction as
\[a + b/c + d\]These students know that they are writing down the original fraction. However, almost anyone else will see the following
\[a + \frac{b}{c} + d\]This is definitely NOT the original fraction. So, if you MUST use “/” to denote fractions use parenthesis to make it clear what is the numerator and what is the denominator. So, you should write it as
\[\left(a + b \right) / \left( c + d \right)\]