# Am I too formal teaching this middle school student?

Solve for $$x$$ the equation below, writing the universe set and the solution set:

$$\frac{a}{b}+\frac{b}{a}-\bigg(\frac{a^2}{bx}+\frac{b^2}{ax}\bigg)=1$$

Let's choose $$U=\{x\in \mathbb R;x\neq 0\}$$ as the universe set.

Using this universe set, the equation above is equivalent to $$(a^2+b^2-ab)x=a^3+b^3$$ which is the same as

$$(a^2+b^2-ab)x=(a+b)(a^2+b^2-ab)$$

Therefore

if $$a^2+b^2-ab=0$$, the solution set is $$S=U$$

If if $$a^2+b^2-ab\neq0$$, the solution set is $$S=\{a+b\}$$

I was teaching my niece in this way, and she said that her teacher said this question has only one solution and it was simpler than mine (she didn't remember very well how he solved).

Maybe, the teacher didn't mention that $$a^2+b^2-ab$$ could be zero, since the school teachers in my home country aren't very formal and sometimes they didn't know how to solve correctly even simple questions.

In my opinion if he solved like this, he is wrong. So my question is how should I teach her? forgetting or pointing out these "small" details? am I too formal and rigorous?

Remark

The real problem is I'm afraid she gets boring and thinks that mathematics is very complicated. Besides that maybe some other teacher or student gives her a simpler wrong solution and finds me a bad teacher who makes things worse. (she doesn't have the mathematical maturity to know which solution is right or wrong).

• Perhaps this is a tangential comment, but is the term "universe set" standard? I think it would be more clear to just say "Find all real numbers $x$ that satisfy the following equation." – littleO Sep 15 '17 at 20:00
• This looks about right. Please don't follow the low standards of english-speaking world in teaching mathematics to pupils. There's a big difference between how you teach and what you teach. What you are teaching looks about right, both in content and in depth. I recall doing thinks like that and even much harder in middle school (continental EU). The difference is in how you deliver this explaination to the students. The pauses you make, the extra reasoning that you do, etc. With the right delivery you can teach MIT undergrad maths to middle school students – Euler_Salter Sep 15 '17 at 20:01
• Also, if you are going to deal with all possible cases, you should also consider the possibility that $a = b = 0$. If this were the case, then you would have $a^2 + b^2 - ab = 0$, but can rule out this case from the original equation for the same reason that we must assume that $x\ne 0$. – Xander Henderson Sep 15 '17 at 20:09
• It is a bit formal for a middle school-er. However $(a^2 - ab + b^2)=0 \iff a=b=0$ in which case the problem is not defined. – Doug M Sep 15 '17 at 20:09
• 1) Since $a, b, x$ appear in denominator, one can implicitly assume $a, b, x \ne 0$, then $a^2 - ab + b^2 = (a-\frac12 b)^2 + \frac34 b^2 \ge \frac34 b^2 > 0$. The teacher is right that this question has only one answer (whether he/she justify $a^2 - ab + b^2 \ne 0$ is another matter). 2) For high school teaching, it is a bad idea to be very formal. The math itself already scaring away a lot of student. If you insists on too formal before they get interested in math, you lose the student forever. – achille hui Sep 15 '17 at 20:41

It is not incorrect to think through all of the cases, and it is not unreasonable to make even a middle schooler think through the possibilities, but you do have to be careful about it, and keep track of the assumptions that have been made. If you are working over the real numbers, then the set of all possible solutions is $\mathscr{U} = \mathbb{R}\setminus \{0\}$, as you noted. However, you seem to be choosing this universe---it is not made explicit in the problem. If you were to choose a different universal set (say $\mathscr{U} = \mathbb{C}\setminus \{0\}$, or $\mathscr{U} = \mathbb{N}$), then the results would be different).

Now, once you have fixed your universe, you are correct that you need to consider the two cases: either (1) $a^2 - ab + b^2 \ne 0$, or (2) $a^2 - ab + b^2 = 0$. Your analysis of case (1) is essentially correct, but your analysis of case (2) is incomplete. To wit:

If $a^2 - ab + b^2 = 0$, then we have \begin{align} &a^2 - ab + b^2 = \left(a-\frac{b}{2}\right)^2 + \frac{3}{4} b^2 = 0 \\ &\qquad \implies \left(a - \frac{b}{4}\right)^2 = -\frac{3}{4}b^2 \\ &\qquad \implies a - \frac{b}{4} = \sqrt{-\frac{3}{4}b^2}. \end{align} If we are assuming that $a$ and $b$ are real (which is something that we have not explicitly done at any point, but our universe of solutions is real, so we have probably done this implicitly), then $b^2$ is nonnegative, and so the quantity on the right will either be pure imaginary, or zero.

If the expression on the right is zero, then $a=b=0$. But if this is so, then the original equation is undefined. There are two possible conclusions: either we live in a universe where such a choice is not possible (i.e. we impose the additional hypothesis that at least one of $a$ and $b$ is nonzero), or the equation is vacuously solved by all possible values of $x$, since a false statement (i.e. $1 = \text{something undefined}$) implies any conclusion.

On the other hand, if the expression on the right is pure imaginary, then we run into other problems, as it would imply that at least one of $a$ and $b$ must be a non-real complex number. Do we allow $a$ and $b$ to be complex, or must they be real? We've made no assumption about what $a$ and $b$ are, but if we want to start with the assumption that the universe of possible $x$ is real, then I think it is safe to assume that $a$ and $b$ are real. In this case, we never conclude that $S = \mathscr{U}$, as we cannot find $a,b\in\mathbb{R}$ such that $a^2-ab+b^2=0$.

Since this is now on the Math Ed SE, let me expand on the first paragraph a bit. Personally, I think that it is important to be precise when doing mathematics, and doubly important when teaching mathematics. I don't think that it is appropriate to tell little children that you can't subtract 5 from 3 (you can, you just get a negative number; better to say "We don't know how yet."), nor do I think that it is okay to say that $x^2 + 2 = 0$ has no solutions (it does, it just doesn't have any real solutions).

Similarly, I think it is wholly appropriate to work through the problem presented above in detail with a middle schooler, particularly if that middle schooler already has enough set theory under their belt to be worried about a universal set of possible solutions.

That being said, I would not present the solution $S = \mathscr{U}$ without also being very careful about determining when this happens. You can't stop at $a^2 - ab + b^2 = 0$. You have to explore what it would mean for this equation to be true, and note that it typically will lead to some kind of contradiction to the assumptions that have already been made.

• Thank you for your answer, the real problem is I'm afraid she gets boring and thinks that mathematics is very complicated. Besides that maybe some other teacher or student gives her a simpler wrong solution and finds me a bad teacher who makes things worse. (she doesn't have the mathematical maturity to know which solution is right or wrong). – user26832 Sep 17 '17 at 17:04
• The edit that you rejected corrects a serious error in the formula. Please be more careful when reviewing edits. – Bill Dubuque Jul 9 '18 at 15:24

Yes. You are messing up now. You can change though.

You should teach the basics first and the caveats, concerns, specialties last.

Teach the kid how to isolate x and get an answer. And THEN double check for some divide by zero concerns.

I think you have to realize that to you the "solve for x" aspect is trivial, while the "watch out for divide by zero" is interesting. But for someone who is new to algebra, the "solve for x" is still new and interesting on its own. [I even feel stronger and think it is the main thing to learn.]

If you do change, it will help the kid and she will enjoy her time with you more also. I think you will like that human interaction in the end also. Give it a shot.

$$a^2-ab+b^2=\frac{3}{4}a^2+\left(\frac{a}{2}-b\right)^2.$$ Thus, if $a^2-ab+b^2=0$ then $a=0$ and $\frac{b}{2}-a=0$, which is impossible.

Hence, $a^2-ab+b^2\neq0$ and we can divide both sides by $a^2-ab+b^2$.

There is an easier way.

Since our equation has unique root (if it has root) and for $x=a+b$ we obtain $$\frac{a}{b}+\frac{b}{a}-\frac{\frac{a^2}{b}+\frac{b^2}{a}}{a+b}-1=\frac{a}{b}+\frac{b}{a}-\frac{\frac{a^3+b^3}{ab}}{a+b}-1=$$ $$=\frac{a}{b}+\frac{b}{a}-\frac{a^2-ab+b^2}{ab}-1=0,$$ we see that $a+b$ is the root.

But the domain gives that $x\neq0$, which gives the answer:

If $a=0$ or $b=0$ or $a=-b$ our equation has no solutions.

If $a\neq0$, $b\neq0$ and $a\neq-b$ then $\{a+b\}$.

You wrote: if a2+b22−ab=0, the solution set is S=U

But your condition cannot be true, which was explained in two ways, in one comment and one answer. It makes some sense to point out that a real mathematician checks this. But the teacher's answer is correct.

• (putting my comment here because I suspect it'll get lost with all the others under the question itself) Personally, I would probably consider solving the special case $a= b = 1$ first, followed by the special case $a = 1$ and $b = 2,$ before dealing with the general case. And it might just be me, but my feeling is that any middle school student who can handle this should be well into graduate level mathematics before finishing high school! – Dave L Renfro Sep 19 '17 at 20:19