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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.

  • 0
    Pl$e$ase 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
  • $x \leq -1$
  • 0
    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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    Nitpick: The number $2$ is not contained in any of your intervals now.2012-07-07