<s>[INST] Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$ My question is: Solve  $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$
I deduced that:$LHS= x+1-(x-2)$
I am unable to solve this equation.  I would like to get some hints to solve it.
 [/INST] $\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|=3$
$|x+1|-|x-2|=3$
1) $x\in(-\infty, -1)$$\Rightarrow$$|x+1|=-(x+1)=-x-1$, $|x-2|=-(x-2)=2-x$.
$|x+1|-|x-2|=3$$\Rightarrow$ $-x-1-2+x=3$$\Rightarrow$$-3=3$, this is a contradiction.
In this interval equation has no solution.
2) $x\in[-1, 2)$$\Rightarrow$ $|x+1|=x+1$, $|x-2|=-(x-2)=2-x$.
$|x+1|-|x-2|=3$$\Rightarrow$ $ x+1-2+x=3$ $\Rightarrow$$2x=4$$\Rightarrow$$x=2$.
$2\notin [-1, 2)$. Also in this interval equation has no solution.
3)  $x\in(2, \infty)$$\Rightarrow$ $|x+1|=x+1$, $|x-2|=x-2$.
$|x+1|-|x-2|=3$$\Rightarrow$ $ x+1-x+2=3$ $\Rightarrow$$3=3$.
On this interval equation has infinity solutions.
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