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In Algebra, a good rule-of-thumb I saw was If solving an equation leads to a contradiction, there is no solution. And this makes sense to me, particularly in the following case:

$x-1 = x +1 \Rightarrow -1 = 1$

We are being asked to find a number whose predecessor is the same as its successor. That would require 1 and -1 to have the same effect on $x$. Impossible! But I was shown another case of no solution -- an equation that seemed to have a valid solution up until checking:

$\frac{2}{x+2} + \frac{x}{x-2} = \frac{x^2+4}{x^2-4} \Rightarrow 2(x-2)+x(x+2) = x^2+4$

Solving this for $x$, we can get an answer of $x=2$. But we see in the original equation, $x\neq2$ because of division by zero. So there is no solution.

My question is: What does it mean (geometrically / algebraically / logically) if solving an equation does not halt in a contradiction, but causes a contradiction on evaluation?

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    The way I understand this is: You multiplied by $(x+2)(x-2)$, which when $x = \pm2$ equals zero, so you basically multiplied the equation by 0, which introduced new solutions to the equation.

    Just like when you take the equation: $x = 5$ and multiply it by 0, you get: $0 = 0$ which means the equation is true for all x, which obviously it isn't.

    Here's a wikipedia article: http://en.wikipedia.org/wiki/Extraneous_and_missing_solutions

     – omer Jul 12 '13 at 23:49

2 Answers2

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$$\frac{2}{x+2} + \frac{x}{x-2} = \frac{x^2+4}{x^2-4} \quad \overset{\large x\neq \pm 2}{\Longrightarrow} \quad2(x-2)+x(x+2) = x^2+4$$

In moving from the left-hand side to the right-hand side of the "implication" above, you implicitly multiplied both terms on the left, along with the right hand side of the equation to the left, by $$x^2 - 4 = (x + 2)(x - 2)$$. If we do not qualify that "operation" by the provision $x\neq \pm 2$, we are allowing for the possibility of multiplication by zero. Any equation, when multiplied by zero, may change in ways that introduce extraneous solutions which aren't actually solutions to the original equation. Without omitting $x = \pm 2$ (to disallow multiplication by zero), we dilute the original equation and information about the original equation is lost.

So the apparent solution of $x = 2$ was introduced by manipulations on the original equation, and in particular, by forgetting the provision I've made explicit above the "implication" connective above. I wouldn't necessarily say that we arrived at a contradiction, but rather found information that turned out to be information (an apparent solution) that doesn't apply to the original equation which has no solution.

amWhy
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  • Including where the expression turns to zero above the arrow makes a lot of sense to me! Thanks! –  Jul 13 '13 at 04:41
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Generally it means that the original function is not defined at the value you arrived at. Usually there is an asymptote, but there can also just be a discontinuity.

Ovi
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  • It's not a function being evaluated, but an equation, and an equation which has no solution: no value x at which equality holds. – amWhy Jul 13 '13 at 00:22
  • @amWhy I know it is not a function, but I think making this problem as finding the zeroes $f(x)=\frac{2}{x+2} + \frac{x}{x-2} - \frac{x^2+4}{x^2-4} $ can give the OP the geometric representation he/she was probably looking for. – Ovi Jul 13 '13 at 00:47