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 04 '12 at 20:58
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Yes i meant the square root – mgh Jun 04 '12 at 21:00
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like the one posted by André Nicolas – mgh Jun 04 '12 at 21:01
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Abstract duplicate of this recent question and this one. – Bill Dubuque Jun 04 '12 at 21:17
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Please refer to : http://math.stackexchange.com/questions/167087/what-is-the-algorithm-for-solving-an-equation-like-this-one – lab bhattacharjee Jul 07 '12 at 12:26
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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:
- $x \geq 2$
- $-1<x<2$
- $x \leq -1$
Gigili
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I didn't write the full solution as you asked for hints to solve it on your own. – Gigili Jun 04 '12 at 20:49
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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.
Cameron Buie
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Madrit Zhaku
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1More 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 07 '12 at 12:10
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