Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3 - Math Stack Exchange

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Learn more about Teams Solving $\sqrt{x^2 +2x + 1}-\sqrt{x^2-4x+4}=3$ Ask Question Asked 11 years, 11 months ago Modified 11 years, 10 months ago Viewed 1k times 4 $\begingroup$

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.

• algebra-precalculus

Share Cite Follow edited Jun 4, 2012 at 21:08 Asaf Karagila♦ 395k4646 gold badges614614 silver badges1k1k bronze badges asked Jun 4, 2012 at 20:41 mghmgh 1,00522 gold badges1010 silver badges2222 bronze badges $\endgroup$ 5

• $\begingroup$ By "whole root" do you mean square root, as in $\sqrt{x^2+2x+1}-\sqrt{x^2-4x+4}=3$? $\endgroup$ – André Nicolas Jun 4, 2012 at 20:58
• $\begingroup$ Yes i meant the square root $\endgroup$ – mgh Jun 4, 2012 at 21:00
• $\begingroup$ like the one posted by André Nicolas $\endgroup$ – mgh Jun 4, 2012 at 21:01
• $\begingroup$ Abstract duplicate of this recent question and this one. $\endgroup$ – Bill Dubuque Jun 4, 2012 at 21:17
• $\begingroup$ Please refer to : math.stackexchange.com/questions/167087/… $\endgroup$ – lab bhattacharjee Jul 7, 2012 at 12:26

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2 Answers 2

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 8 $\begingroup$

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

You have to consider three cases:

• $x \geq 2$
• $-1<x<2$
• $x \leq -1$

Share Cite Follow edited Jun 4, 2012 at 21:01 answered Jun 4, 2012 at 20:45 GigiliGigili 5,51388 gold badges4141 silver badges6262 bronze badges $\endgroup$ 3

• $\begingroup$ I didn't write the full solution as you asked for hints to solve it on your own. $\endgroup$ – Gigili Jun 4, 2012 at 20:49
• $\begingroup$ Hey but there is a whole root sign $\endgroup$ – mgh Jun 4, 2012 at 20:55
• $\begingroup$ Oops, sorry. I'm not used to English-mathematics-nolatex notation! $\endgroup$ – Gigili Jun 4, 2012 at 20:57

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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.

Share Cite Follow edited Jul 7, 2012 at 12:06 Cameron Buie 104k99 gold badges103103 silver badges226226 bronze badges answered Jul 7, 2012 at 11:40 Madrit ZhakuMadrit Zhaku 5,30433 gold badges1616 silver badges1919 bronze badges $\endgroup$ 2

• 1 $\begingroup$ 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 $2$. $\endgroup$ – Cameron Buie Jul 7, 2012 at 12:10
• $\begingroup$ Nitpick: The number $2$ is not contained in any of your intervals now. $\endgroup$ – TMM Jul 7, 2012 at 13:04

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