# Thread: Inequality with absolute values

1. ## Inequality with absolute values

Solve for
|x(x+1)| < |x +4|

[x(x+1)] 2 < (x+4)2
[x(x+1)] 2 (x+4)2 < 0
[x (x+1)
(x+4)] [x (x+1) + (x+4)] < 0
(
x2 + x – x – 4) (x2 + x + x + 4) < 0
(x2 4) (x2 + 2x + 4) < 0

the product of two factors is negative since it is less than (<) zero (0).

and
x2 + 2x + 4 = x2 + 2x + 1 + 3
=
(x + 1)2 + 3 then it is positive.

then (x2 4) is negative.
Finally:
x2 4 < 0
x2 < 4
take the sq. root on both sides, then
|x| < 2 Q.E.D

2. [x(x+1)] ^2 < (x+4)^2 this has to be like this... i don't know how to make or raise the 2 as exponent. please understand please help me rewrite the posted and solved problem

[x(x+1)]^ 2 < (x+4)^2
[x(x+1)] ^2 – (x+4)^2 < 0
[x (x+1) – (x+4)] [x (x+1) + (x+4)] < 0
(x^2 + x – x – 4) (x^2 + x + x + 4) < 0
(x^2 – 4) (x^2 + 2x + 4) < 0

the product of two factors is negative since it is less than (<) zero (0).

and x^2 + 2x + 4 = x^2 + 2x + 1 + 3
= (x + 1)^2 + 3 then it is positive.

then (x^2 – 4) is negative.
Finally:
x^2 – 4 < 0
x^2 < 4
take the sq. root on both sides, then
|x| < 2 Q.E.D