Solve
$\displaystyle \sqrt{x^{2}+4x-3}=1-2x.$
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Solve
$\displaystyle \sqrt{x^{2}+4x-3}=1-2x.$
The real "solutions" 2 and 2/3 are both extraneous as 1-2x is negative for both.
That's right. I just thought it was a cute problem, because, although in the process of solving the problem, you never do anything "really dangerous", and you actually come up with formal "solutions", neither of them work. Usually, if you have extraneous solutions to a quadratic-type problem, only one of them will be extraneous, and the other is ok.
Good job.
