How do I solve |x+1| < |x-3|
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 8, 2010 #2 \(\displaystyle |x+1| \ < \ |x-3|, \ one \ way \ squaring \ both \ sides \ (gets \ rid \ of \ the \ pesky \ absolute \ value\) \(\displaystyle signs) \ gives:\) \(\displaystyle |(x+1)^2| \ < \ |(x-3)^2| \ = \ x^2+2x+1 \ < \ x^2-6x+9\) \(\displaystyle \implies \ 8x \ < \ 8 \ \implies \ x \ < \ 1.\)
\(\displaystyle |x+1| \ < \ |x-3|, \ one \ way \ squaring \ both \ sides \ (gets \ rid \ of \ the \ pesky \ absolute \ value\) \(\displaystyle signs) \ gives:\) \(\displaystyle |(x+1)^2| \ < \ |(x-3)^2| \ = \ x^2+2x+1 \ < \ x^2-6x+9\) \(\displaystyle \implies \ 8x \ < \ 8 \ \implies \ x \ < \ 1.\)
