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

*Solve for all real values of :*

We’ve had an active discussion about this problem on their facebook page, and you should go check it out and join the conversation yourself. Go ahead and try it if you haven’t already.

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

Okay, here’s the work. Factor everything.

Multiply both sides by the denominator.

Use the zero-product property to find . Now check for extraneous solutions and find that and give you and gives . This last statement DOES actually hold for but we exclude it because it’s not in the domain of the original expression.The original expression has domain . 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 .

**Does this seem problematic?** How can we exclude 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:

**Same question: Is 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 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 *:

Is a solution? No, I think we must reject it. If we try to check it, we must evaluate but this expression is undefined because . Here’s another one:

*Solve over *:

Is a solution? No. Now when we try to check the solution we get which is undefined.

The point is that, if we go back to the same question and ask about the solutions of over the reals, and we check the solution , we must evaluate which is undefined in the reals.[1]

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

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.

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[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! 🙂