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William from Arizona asks, "Myself and a fellow math teacher of 20+ years each got into a discussion about extraneous solutions for a particular problem. The problem is this: 1/(x-a) = x/(x-a) The provided answer to this problem is that x = a is always an extraneous solution. However, when you solve it using the LCD method to multiply both sides, the x = a solution candidate does not present itself. (x-a) * 1/(x-a) = x/(x-a) *(x-a) 1 = x cancelling x-a The only candidate after cancelling is x = 1. The counter argument was that you solve by cross multiplying and setting equal getting: x - a = x(x - a) x - a = x^2 - ax 0 = x^2 - ax - x - a 0 = x^2 - (a+1)x - a Using quadratic formula: x = a or x = 1 therefore, x = a is extraneous. Using the cross multiply method, I run into x = a and need to call it extraneous. Using LCD multiply to both sides, it doesn't present as a candidate. So, is it true to say x = a is always an extraneous solution? Thank you for considering"

Hi William, thanks for asking. This is an interesting question. An extraneous solution is generally defined as "a solution to a transformed equation which is not a solution of the original equation." So, for example, if you square both sides of an equation, you've transformed it, and therefore you've introduced the possibility of solutions that might not be solutions to the original equation.

In this case, the transformation that produces an extraneous solution is cross multiplying, which removes the (x - a) from the denominator (and therefore including x = a in the domain).

So an extraneous solution is a solution that arises because of a transformation. If the transformations you performed did not result in a new equation that has an expanded domain, then it doesn't produce extraneous solutions. In other words, no, I don't think you have an extraneous solution using the first method.

The phrasing here is odd; if x = 1 was an extraneous solution, you wouldn't say "x = 1 is always an extraneous solution." I suspect that what they're trying to say is, "it doesn't matter what the value of 'a' is; x = a will never be a solution." So I don't think they were trying to suggest that there's no way to avoid x = a as an extraneous solution, but rather that if it does arise as a solution, it is always extraneous, regardless of the value of 'a'.

As one last note, I'd like to point out that you can avoid using the quadratic formula in your second method:

(x - a) = x(x - a)
0 = x(x - a) - (x - a)  //subtract (x - a) from both sides
0 = (x - a)(x - 1) //use the distributive property to rewrite the previous equation
x =a or x = 1