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Solve $\mid x^2+4x+3 \mid +~2x+5=0$

My work:
I know this is a simple problem and I can solve it by means of going-on and just keep on manipulating each and every step. But, I want to know are there no other better way by which I can reach the solution without much effort and which is quite elegant and requires thinking skills.

EDIT: I removed "robotic" as it stands to be objectionable by DonAntonio :) and since I do not know about robotics and he knows so I remove it.

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  • $\begingroup$ Please explain how you normally would do it. $\endgroup$
    – naslundx
    Feb 26, 2014 at 12:15
  • $\begingroup$ @naslundx First removing the modulus and then simple manipulations and factoring. $\endgroup$
    – Hawk
    Feb 26, 2014 at 12:17
  • $\begingroup$ What is $x$? A real number? An integer? $\endgroup$
    – frabala
    Feb 26, 2014 at 12:18
  • $\begingroup$ "Robotic going on"?? I can't see how robots get into this. In fact, it requires some relevant thinking imo. $\endgroup$
    – DonAntonio
    Feb 26, 2014 at 12:20

2 Answers 2

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Hints:

$$x^2+4x+3=(x+1)(x+3)\ge0\iff\begin{cases}x\le -3\;\;or\\x\ge-1\end{cases}$$

and thus you have to solve the system

$$\begin{cases}0=x^2+4x+3+2x+5=x^2+6x+8&,\;\;x\le-3\;\;or\;\;x\ge-1\\{}\\0=-x^2-4x-3+2x+5\iff x^2+2x-2=0&,\;\;-3<x<-1\end{cases}$$

Both cases are pretty straightforward...

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  • $\begingroup$ Yes, I know this solution. But, this doesn't look very elegant. That is why I asked for something new and something elegant. $\endgroup$
    – Hawk
    Feb 26, 2014 at 12:25
  • $\begingroup$ Personally I would first try to solve both equations and then go back and check if the answer checks with the sign I assumed to the modulus. But this is a matter of personal preference... $\endgroup$
    – lsoranco
    Feb 26, 2014 at 12:27
  • $\begingroup$ "Doesn't look very elegant"?? Well, I strongly beg to differ: it proves knowledge of what (real) absolute value actually means, and requires to solve both quadratic equations and a system of two such, and it divides the problem in two clear, simpler cases. Of course, in such a basic problem the possibilities are not many to play with possible ways for solution, but you can try... $\endgroup$
    – DonAntonio
    Feb 26, 2014 at 12:27
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    $\begingroup$ It never matters, @Zado: anyway you must know where the expression whithin absolute value is positive and where not... $\endgroup$
    – DonAntonio
    Feb 26, 2014 at 12:28
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Let $a=x+1$ and $b=x+3$, thus we have to solve $|a|\cdot|b|+a+b+1=0$. Iff $x\ge-1$ or $x\le-3$ this equation becomes $ab+a+b+1=0\iff (a+1)(b+1)=0$. Now iff $-3<x<-1$ the original equation has exactly one solution.

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