absolute value inequalities w/ abs val exp on both sides

x+1 < x3, one way squaring both sides (gets rid of the pesky absolute value\displaystyle |x+1| \ < \ |x-3|, \ one \ way \ squaring \ both \ sides \ (gets \ rid \ of \ the \ pesky \ absolute \ value

signs) gives:\displaystyle signs) \ gives:

(x+1)2 < (x3)2 = x2+2x+1 < x26x+9\displaystyle |(x+1)^2| \ < \ |(x-3)^2| \ = \ x^2+2x+1 \ < \ x^2-6x+9

     8x < 8      x < 1.\displaystyle \implies \ 8x \ < \ 8 \ \implies \ x \ < \ 1.
 
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