I looked at a video from Khan Academy on limits at infinity and in the video, the instructor said $\sqrt {x^2}$ is 'essentially taking the absolute value of x; |x|'. I couldn't understand why is the same thing. Is it because when we take the principal root of $x^2$, we are always going to have a positive outcome and absolute value does that as well? So, Why and When do we use the absolute value?

Thank you. P.S I am just learning the basics of calculus.

  • $\begingroup$ $|x|$ is nice to write than $\sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses. $\endgroup$ – Zachary Hunter Jan 6 at 7:06
  • $\begingroup$ @ZacharyHunter they are the samething? So my assumption is correct? $\endgroup$ – Fred Weasley Jan 6 at 7:08
  • $\begingroup$ Yes they are the same, $|x|$ is just more concise and all. $\endgroup$ – Zachary Hunter Jan 6 at 7:09
  • $\begingroup$ Yes, your reason is correct. $\sqrt{x^2}=|x|\ne x\forall x <0$; for example, $\sqrt{(-5)^2}=5=|-5|$ $\endgroup$ – Shubham Johri Jan 6 at 7:28
  • $\begingroup$ Just a small nitpick: $\sqrt{x^2} = \vert x\vert$ means the outcome is always non-negative since $x$ can be $0$. $\endgroup$ – KM101 Jan 6 at 7:34

Yes, $\vert x\vert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:

$$\sqrt{\color{blue}{+4}^2} = \sqrt{16} = \color{blue}{+4}$$

$$\sqrt{\color{blue}{+9}^2} = \sqrt{81} = \color{blue}{+9}$$

$$\sqrt{(\color{blue}{-4}^2)} = \sqrt{16} = \color{blue}{+4} = -(\color{blue}{-4})$$

$$\sqrt{(\color{blue}{-9}^2)} = \sqrt{81} = \color{blue}{+9} = -(\color{blue}{-9})$$

So, when $x > 0$, then $\sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $\sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,

$$\sqrt{x^2} = \vert x\vert = \begin{cases} x; \quad x \geq 0 \\ -x; \quad x < 0 \end{cases}$$

This is all just a way of saying $\sqrt{x^2}$ will always return a non-negative value.

  • $\begingroup$ Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics? $\endgroup$ – Fred Weasley Jan 6 at 8:15
  • $\begingroup$ It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself. $\endgroup$ – KM101 Jan 6 at 8:45

I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.

But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.


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