Question

The solutions set of $\left|\frac{x+1}{x}\right|+x+1=\frac{(x+1{)}^2}{|x|}$ is

• A. $x|x\ge 0$
• B. $x|x>0\cup -1$
• C. $-1,1$
• D. $x|x\ge 1orx\le -1$

Answer 1

The correct option is B $x|x>0\cup -1$

$\left|\frac{x+1}{x}\right|+(x+1)=\frac{(x+1{)}^2}{|x|}$

Case 1 If $x\le -1\Rightarrow \frac{x+1}{x}>0$ and $|x|=-x$

$\frac{x+1}{x}+(x+1)=\frac{(x+1{)}^2}{-x}$, given equation becomes

$\Rightarrow \frac{(x+1{)}^2}{x}=\frac{(x+1{)}^2}{-x}\Rightarrow x=-1$

Case 2 If $-1<x<0\Rightarrow \frac{x+1}{x}<0$ and $|x|=-x$

$-\frac{x+1}{x}+(x+1)=\frac{(x+1{)}^2}{-x}$, given equation becomes

$\Rightarrow \frac{(x+1)(x-1)}{x}=\frac{(x+1{)}^2}{-x}\Rightarrow$ no solution

Case 3 If $x>1\Rightarrow \frac{x+1}{x}>0$ and $|x|=x$

$\frac{x+1}{x}+(x+1)=\frac{(x+1{)}^2}{x}$, given equation becomes

$\Rightarrow \frac{(x+1{)}^2}{x}=\frac{(x+1{)}^2}{x}\Rightarrow \forall x>0$

Hence solution is $x|x>0\cup \{-1\}$