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Old 12-31-2007   #1
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Default surd inequality

please help with it:

square root(x+2-x^2)<square root(x^2-3x+2)+x

Last edited by agripa; 01-01-2008 at 10:23 AM..
Old 01-01-2008   #2
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Default be careful

Square roots provide plenty of opportunities to make mistakes. So do inequalities.
There may be a much easier way that I don't see, but I would square.
If 0<a<b, then a^2<b^2
Since x+2+x^2 = (x+1/2)^2 + 7/4 > 0 , you can square to get
-x^2+4x < 2xsqrt(x^2-3x+2)
After that you have to consider cases (x<0, x>0, x=4, x>4, x<4, etc) and square again.

Happy New Year!

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