# Great NCTM problem

Yesterday I presented this problem from NCTM’s facebook page:

Solve for all real values of $x$:

$\frac{(x^2-13x+40)(x^2-13x+42)}{\sqrt{x^2-12x+35}}$

Don’t read below until you’ve tried it for yourself.

Okay, here’s the work. Factor everything.

$\frac{(x-8)(x-5)(x-7)(x-6)}{\sqrt{(x-5)(x-7)}}=0$

Multiply both sides by the denominator.

$(x-8)(x-5)(x-7)(x-6)=0$

Use the zero-product property to find $x=5,6,7,8$. Now check for extraneous solutions and find that $x=5$ and $x=7$ give you $\frac{0}{0}\neq 0$ and $x=6$ gives $x=\frac{0}{\sqrt{-1}}=\frac{0}{i}=0$. This last statement DOES actually hold for $x=6$ but we exclude it because it’s not in the domain of the original expression.The original expression has domain $(-\infty,5)\cup(7,\infty)$. We could have started by identifying this, and right away we would know not to give any solutions outside this domain. The only solution is $x=8$.

Does this seem problematic? How can we exclude $x=6$ as a solution when it (a) satisfies the equation and (b) is a real solution? This is why we had such a lively discussion.

But this equation could be replaced with a simpler equation. Here’s one that raises the same issue:

Solve for all real values of x:

$\frac{x+5}{\sqrt{x}}=0$

Same question: Is $x=-5$ a solution? Again, notice that it DOES satisfy the equation and it IS a real solution. So why would we exclude it?

Of course a line is drawn in the sand and many people fall on one side and many fall on the other. It’s my impression that high-school math curriculum/textbooks would exclude $x=-5$ as a solution.

Here’s the big question: What does it mean to “solve for all real values of x“? Let’s consider the above equation within some other contexts:

Solve over $\mathbb{Z}$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No, I think we must reject it. If we try to check it, we must evaluate $\frac{0}{\sqrt{5}}$ but this expression is undefined because $\sqrt{5}\notin\mathbb{Z}$. Here’s another one:

Solve over $\mathbb{Z}_5$:

$\frac{x+5}{\sqrt{x}}=0$

Is $x=-5$ a solution? No. Now when we try to check the solution we get $\frac{0}{\sqrt{5}}=\frac{0}{\sqrt{0}}=\frac{0}{0}$ which is undefined.

The point is that, if we go back to the same question and ask about the solutions of $\frac{x+5}{\sqrt{x}}=0$ over the reals, and we check the solution $x=-5$, we must evaluate $\frac{0}{\sqrt{-5}}$ which is undefined in the reals.[1]

So in the original NCTM question, we must exclude $x=6$ for the same reason. When you test this value, you get $\frac{0}{i}$ on the left side which YOU may think is 0. But this is news to the real numbers. The reals have no idea what $\frac{0}{i}$ evaluates to. It may as well be $\frac{0}{\text{moose}}$.

There’s a lot more to say here, so perhaps I’ll return to this topic another time. Special thanks to all the other folks on facebook who contributed to the discussion, especially my dad who helped me sort some of this out. Feel free to comment below, even if it means bringing a contrary viewpoint to the table.

________________________

[1] This last bit of work, where we fix the equation and change the domain of interest touches on the mathematical concept of algebraic varieties, which I claim to know *nothing* about. If someone comes across this post who can help us out, I’d be grateful! 🙂

## 7 thoughts on “Great NCTM problem”

1. -5 isn’t in Z_5, or rather it’s already 0 in Z_5, but we purposely confuse equivalence classes and representatives of the classes all the time, so No Big Deal: -5 = 5 = 0 in Z_5.

My Big Deal is why the Wolfram Alpha site refuses to see all of these obvious things that we’re talking about. It will make the domain of a function on Reals as large as possible whether we want it to or not. But then it doesn’t go far enough. It should allow us to say, “Please solve this in the Reals.” It did do this for me correctly:
solve 3x^2+x-7=4x in Z_11

Stepping out of a system and landing back in it reminds me of this following very mysterious fact that no mathematician understands. There are facts about integers whose only known proof uses analysis, often complex analysis. Is there a (meta-) theorem that guarantees that any proof of a statement about integers can be proven using only facts about integers, and not complex numbers? We call the first kind of proving doing “analytic number theory”; we call the second kind of proving “elementary number theory.” Elementary number theory is harder than analytic number theory, which shows you that sometimes going to a more encompassing system actually simplifies a problem. My intuition is that the statement “every theorem of analytic number theory is a theorem of elementary number theory” is unprovable, as is its negation.

2. As I posted to the NCTM Facebook page, I contend that x=5, x=6, x=7, and x=8 are all correct solutions to this problem. The equation can be simplified using algebra to (x-8)*sqrt(x-5)*sqrt(x-7)*(x-6)=0. For all real values of x, there are no indeterminate values. Or do you contend that (y*y)/y=0 –> y=0 is undefined at y=0?

• Right. y*y/y is undefined at zero, if we’re working in a field like the reals or complex numbers. More explicitly, 0/0 is undefined. Wouldn’t you agree that 0/0 is undefined? Check it on a calculator, too!

• Of course, 0/0 is undefined. According to Mathematica (a pretty decent calculator), y=0 *is* a solution:

Solve[y^2/y==0,y]
{{y->0}}

3. Same results even when we explicitly tell Mathematica that y is an element of the reals:

Solve[Assuming[y ∈ Reals, y^2/y] == 0, y]
{{y -> 0}}

4. Hmm.. that’s really interesting. After poking around a bit, I think this is an error in Mathematica/Wolfram Alpha.

If I ask for the value of 0/0, it correctly says “indeterminate”:
http://www.wolframalpha.com/input/?i=0%2F0

If I ask for the value of x/x evaluated at x=0, it says “1”, which I consider to be incorrect:
http://www.wolframalpha.com/input/?i=x%2Fx+evaluate+when+x%3D0

There’s a short discussion of this bug here:
http://community.wolframalpha.com/viewtopic.php?f=32&t=70878

Let me know if you discover anything else. I’m interested.