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Are my questions invalid or difficult cause I'm not getting answers since many days?

Question 1:
     By definition absolute value gives just no of units and does not indicate any direction neither positive nor negative then why in practice we use +ve direction like $\left|4\right|=+4$ it should be just 4 not +4

Question 2:
     We know that   $\sqrt{{x}^2} = \pm x = \left|x\right| $  

Then why   $\left|4\right|=+4$   and not   $\left|4\right| = \pm 4$
Also     $\sqrt{{4}^2} = +4$   and not   $\sqrt{{4}^2} = \pm 4$

Is this all have do with involvement of a variable only (when variable involved use $\pm$ and when constants involved don't use $\pm$)   ............ or there are some other rules !

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The plus does nothing, it's $+4=4$. It's there to draw attention to the fact that the absolute value left a positive value unchanged -- they wanted to stress the difference between $-$ and $+$ case.

Also, $\sqrt{x^2}=\pm x$ is wrong. It's $\sqrt{x^2}=|x|$. The result of square root is always positive by definition. You can't get two values from a function. If you want to solve a quadratic equation, for instance, $y^2=x$, then you have to explicitly put $\pm$ in front of the square root to specify both solutions: $y=\pm \sqrt{x}$.

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  • $\begingroup$ Is this true, $$ \left|k\right| < x \implies -k < x < +k $$ $\endgroup$ – Danish ALI Nov 15 '14 at 19:55
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    $\begingroup$ @DanishALI No, it's the other way around. If $x>0$, then you just bounded $k$ from both sides, so $-x<k<x$ (you can interpret this statement as "the distance of k from the coordinate origin is less than x". However, if $x\leq 0$, then no $k$ satisfies this condition (the absolute value cannot be negative). $\endgroup$ – orion Nov 16 '14 at 9:52
  • $\begingroup$ I have a link to this question: math.stackexchange.com/questions/96065/… here and in this post a reply by David Mitra states that |x + 1| is either (x + 1) or -(x + 1) if we interpret it here we could say |x| is either x or -x so is David Mitra saying wrong ? because |x| can't be -x because absolute value function will always produce positive numbers $\endgroup$ – Danish ALI Nov 16 '14 at 15:49
  • $\begingroup$ @DanishALI You are understanding it wrong. $|x|$ is one of the options $x$ and $-x$: the one that makes the value positive! It's not both and it's not optional what you choose! So, $|-5|=-(-5)$ because this makes it positive, but $|5|=(5)$ because that stays positive. $\endgroup$ – orion Nov 16 '14 at 16:21
  • $\begingroup$ Alright I get it this way |x| will always be positive but x could be any positive number or a negative number but when x goes under || operator it's returned in positive regardless what original direction was. Now have I understood it correctly ? $\endgroup$ – Danish ALI Nov 17 '14 at 20:00

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