Question

Number of integral solutions of $|x^2+4x+4|-|2x+4|+1<\frac{1}{2}$ is

Answer 1

$|x^2+4x+4|-|2x+4|+1<\frac{1}{2}\Rightarrow (|x+2|-1{)}^2<\frac{1}{2}\Rightarrow -\frac{1}{\sqrt{2}}<|x+2|-1<\frac{1}{\sqrt{2}}\Rightarrow -\frac{1}{\sqrt{2}}+1<|x+2|<1+\frac{1}{\sqrt{2}}$

If $x\ge -2$,
then $-\frac{1}{\sqrt{2}}-1<x<\frac{1}{\sqrt{2}}-1$
Possible integral value of $x$ is $1$
Number of integral solutions is $1$

If $x<-2$,
then $1-\frac{1}{\sqrt{2}}<-x-2<1+\frac{1}{\sqrt{2}}$
$\Rightarrow -3+\frac{1}{\sqrt{2}}>x>-3-\frac{1}{\sqrt{2}}$
Possible integral value of $x$ is $-3$
Number of integral solutions is $1$

$\therefore$ Total number of integral solutions is $2.$