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01012008  #1 
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algebra need help
\sqrt {x + 2  x^2 } < \sqrt {x^2  3x + 2} + x. please help me

01012008  #2 
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Not as hard as it seems
I cannot get the symbols right either.
For the square roots to be real numbers, x^2+x+2 and x^23x+2 must be positive or zero That only happens for x=1, x=1 and all numbers in between. That is the domain you are working in. Luckily for you, neither side of the inequality is negative there, so you can square both sides to get another inequality with exactly the same solutions. Rearrange the new inequality so that one side contains only the term with the square root. You'll find x as a factor on both sides. For x=0, both sides of your inequality are equal, so x=0 is not a solution. You will have to consider two cases (x>0 and x<0). Squaring again, you get an inequality that is never true for x>0. For x<0, the last inequality is always true within your domain. Your solution, then, is x=1 and 1<x<0. For your comfort (not a real check), you can try values of x (e.g. x=1, x=1/2, x=0) in the original inequality. 
01022008  #3 
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Why not just for notation?

01022008  #4 
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