- #1

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I end up obtaining two solutions: ##x = 6## and ##x = 22/49##. Plugging these into the original equation, I find that ##x = 6## works, but ##x = 22/49## does not. I understand that the latter is termed an extraneous solution. My question is, how do these extraneous solutions arise? Does it have something to do with the fact that we lose information about the original equation when we square both sides?