Take (x+1)^2 = x^2 + 2x +1 which is true for all x. Then we have x = 2x^2-5x which is true for x = 0 and x = 3. If we have any combination, I think you can only use x which is valid in both, which is just 0 and 3. Then writing (2x^2 - 5x + 1)^2 = x^2 + 2x +1, should only be satisfied by 0 and 3, however this is not the case ( we have an extra solution x=1). Now I would like to ask, does this extra solution affect the logic of what we already did? What I mean by this is that just because we could only be sure that the equation held for some x, does that interfere with the logic that it holds for other x which we did not expect? Or is it a case of, all we can say is that it holds for 0 and 3. (Not that they are the only solutions, but that it definitely holds for those x)
Where did the extra solutions come from?
- Thread starter nav22
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There are no "extra" solutions.
You have merely created a new function from 2 other expressions (one of which just happens to be the definition of the distributive law).
You have merely created a new function from 2 other expressions (one of which just happens to be the definition of the distributive law).
