# Hassle with Absolute Value and Square Root

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 !

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: $y=\pm \sqrt{x}$.
• Is this true, $$\left|k\right| < x \implies -k < x < +k$$ – Danish ALI Nov 15 '14 at 19:55
• @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). – orion Nov 16 '14 at 9:52
• @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. – orion Nov 16 '14 at 16:21