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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 $\sqrt{x^2+2x+1}-\sqrt{x^2-4x+4}=3$?2012-06-04
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    Yes i meant the square root2012-06-04
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    like the one posted by André Nicolas2012-06-04
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    Abstract duplicate of [this recent question](http://math.stackexchange.com/questions/98157) and [this one.](http://math.stackexchange.com/questions/153818)2012-06-04
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    Please refer to : http://math.stackexchange.com/questions/167087/what-is-the-algorithm-for-solving-an-equation-like-this-one2012-07-07

2 Answers 2

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

You have to consider three cases:

  • $x \geq 2$
  • $-1
  • $x \leq -1$
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    I didn't write the full solution as you asked for hints to solve it on your own.2012-06-04
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    Hey but there is a whole root sign2012-06-04
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    Oops, sorry. I'm not used to English-mathematics-nolatex notation!2012-06-04
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$\sqrt {x^2 +2x + 1}-\sqrt { x^2-4x+4}= \sqrt{(x+1)^2} - \sqrt{(x-2)^2}=|x+1|-|x-2|=3$

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

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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    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 $2$.2012-07-07
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    Nitpick: The number $2$ is not contained in any of your intervals now.2012-07-07