Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
By "whole root" do you mean square root, as in $\sqrt{x^2+2x+1}-\sqrt{x^2-4x+4}=3$? – André Nicolas Jun 4 '12 at 20:58
Yes i meant the square root – mgh Jun 4 '12 at 21:00
like the one posted by André Nicolas – mgh Jun 4 '12 at 21:01
Abstract duplicate of this recent question and this one. – Bill Dubuque Jun 4 '12 at 21:17
Please refer to :… – lab bhattacharjee Jul 7 '12 at 12:26

$$\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|$$

You have to consider three cases:

  • $x \geq 2$
  • $-1<x<2$
  • $x \leq -1$
share|cite|improve this answer
I didn't write the full solution as you asked for hints to solve it on your own. – Gigili Jun 4 '12 at 20:49
Hey but there is a whole root sign – mgh Jun 4 '12 at 20:55
Oops, sorry. I'm not used to English-mathematics-nolatex notation! – Gigili Jun 4 '12 at 20:57

$\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|=3$


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.

share|cite|improve this answer
More precisely, the last interval is the solution set. It would still have "infinity solutions" if the solution set were, say, $(2,3)$, or the set of integers greater than $2$. – Cameron Buie Jul 7 '12 at 12:10
Nitpick: The number $2$ is not contained in any of your intervals now. – TMM Jul 7 '12 at 13:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.