[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