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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.

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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 : math.stackexchange.com/questions/167087/… –  lab bhattacharjee Jul 7 '12 at 12:26

2 Answers 2

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

You have to consider three cases:

  • $x \geq 2$
  • $-1<x<2$
  • $x \leq -1$
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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

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

$|x+1|-|x-2|=3$

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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1  
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

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