Hi, Le 16/12/2014 01:20, rjf a écrit :
Well, you could assert that there is no discussion, but you are apparently wrong.
May I suggest you to square the equation then solve the obtained polynomial equation? (Then of course, since squaring is an implication, check that the obtained solutions are solutions of the original equation.)
I don't think any discussion of extending sqrt in any way will change the fact that the above gives {0; 1} as a set of solutions.
The fact that you're the only one having issues with this equation should be a pretty strong indication that you're the one off track!
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