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.
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.
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.
