How do I solve |x+1| < |x-3|
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Sep 8, 2010 #2 ∣x+1∣ < ∣x−3∣, 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∣x+1∣ < ∣x−3∣, one way squaring both sides (gets rid of the pesky absolute value signs) gives:\displaystyle signs) \ gives:signs) gives: ∣(x+1)2∣ < ∣(x−3)2∣ = x2+2x+1 < x2−6x+9\displaystyle |(x+1)^2| \ < \ |(x-3)^2| \ = \ x^2+2x+1 \ < \ x^2-6x+9∣(x+1)2∣ < ∣(x−3)2∣ = x2+2x+1 < x2−6x+9 ⟹ 8x < 8 ⟹ x < 1.\displaystyle \implies \ 8x \ < \ 8 \ \implies \ x \ < \ 1.⟹ 8x < 8 ⟹ x < 1.
∣x+1∣ < ∣x−3∣, 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∣x+1∣ < ∣x−3∣, one way squaring both sides (gets rid of the pesky absolute value signs) gives:\displaystyle signs) \ gives:signs) gives: ∣(x+1)2∣ < ∣(x−3)2∣ = x2+2x+1 < x2−6x+9\displaystyle |(x+1)^2| \ < \ |(x-3)^2| \ = \ x^2+2x+1 \ < \ x^2-6x+9∣(x+1)2∣ < ∣(x−3)2∣ = x2+2x+1 < x2−6x+9 ⟹ 8x < 8 ⟹ x < 1.\displaystyle \implies \ 8x \ < \ 8 \ \implies \ x \ < \ 1.⟹ 8x < 8 ⟹ x < 1.