Some mathematical things that I have wondered about for a long time...occasionally get answered. Maybe, it is because I was not smart enough to figure them out or understand them...or maybe I had just not hit on the right explanation at the right time.
An example is the creation of false or extraneous roots when solving certain forms of equations. I can remember teaching the "concern" in high school...and it somehow surfaced in a recent discussion with a student completing a mathematics history project. And, now I know the answer....thanks to an algebra text from 1826.
Suppose you are to solve this equation: x + sqrt(5x+10) = 8
Rewrite it with the sqrt-term isolated: sqrt(5x+10) = 8 - x
Square both sides to eliminate the sqrt-term:
5x + 10 = 64 - 16x + x2
Collect terms on one side: x2 - 21 x +54 = 0
Factor this quadratic into two linear factors: (x-18)(x-3) = 0
Thus, we have two solutions: x = 18 and x = 3
But, by checking, we learn that x = 18 is an extraneous solution (i.e or false root). And, our only solution is x = 3.
So, your task is: Explain how/why/when the extraneous root of x =18 was created...
Hint: Review each step....a big step is to know "when" the false root was introduced...then you can focus on the why/how.
Solution Commentary: According to Enoch Lewis's Practical Analyst or a Treatise on Algebra (1826): "The ambiguity is introduced into this solution by the involution of the given equation..."
Now, I hear you ask, what does "involution" mean? In the "olde" days, "involution" meant raising to a power and "evolution" meant finding a root.
The key step is involution or squaring of both sides of equation sqrt(5x+10) = 8-x
We would have reached the same result if we had involved or squared both sides of the equation sqrt(5x+10) = x-8
Because, (x-8)2 = (8-x)2.
And by checking, you find x = 18 is a "true" root of the equation sqrt(5x+10) = x - 8 while x = 3 is a "true" root of the equation sqrt(5x+10) = 8 - x.
J.S. (Bellingham) offered these insights: "I always tried to get them to think about it in the following way:
First, I would show a simple equation like X = 2.
Pretty simple ... only one solution. It SAYS, " The quantity 'X' is two."
But if I multiply both sides by "X", and assuming that X has the same value throughout,
then the equation says,
X squared = 2X
or X squared - 2X = 0
(X-2)X = 0
which has two solutions, namely X=2 and X=0. Whereas it is true that X = 2 is still valid, it is nonetheless obvious that there is a problem with X = 0 in the original equation.
The same thing happens if you square both sides, although the results are a little different. Then you get:
X = 2
X squared = 4
which leads to X = 2 and X = -2, and the obvious problem is that "-2" will not work in the original equation. I would then point out that in both situations, the equation was being transformed by multiplication involving a variable. I would then mention that if, while solving an equation, you saw the need to multiply one ore more sides of an equation by a variable, then you MUST check your answer. I didn't feel the need to try to go any further than that, for the purposes of an algebra 1 or 2 class.