Verified answer

All values of [tex]\rm{x}[/tex] in reals satisfy [tex]\sf{\sqrt{x^{2}}=|x|}[/tex].

[tex]\;[/tex]

[tex]\Large\textbf{\underline{Explanation}}[/tex]

The square root function is the inverse function.

[tex]\bigstar\underline{\sf{g(x)=x^{2}\ (x\geq0)\textsf{ then }g^{-1}(x)=\sqrt{x}}}[/tex]

The correlation between the two functions is as follows.

[tex]\bigstar\underline{\sf{(g^{-1}\circ g)(x)=x\ (x\geq0)}}[/tex]

And we know the definition of the absolute value function.

Absolute value function

[tex]{\boxed{\sf |x|=\begin{cases}&\sf x\ (x \geq 0)\\\\&\sf -x\ (x < 0)\end{cases}}}[/tex]

[tex]\;[/tex]

From the correlation and definition, we get the result.

[tex]\Longrightarrow\sf{(g^{-1}\circ g)(x)=|x|\ (x\geq0)}[/tex]

[tex]\Longrightarrow\sf{\sqrt{x^{2}}=|x|\ (x\geq0)}[/tex]

[tex]\;[/tex]

Composite [tex]\sf{-x}[/tex] into the function, so we get the rest of the results.

[tex]\Longrightarrow\sf{\sqrt{(-x)^{2}}=|-x|\ (-x < 0)}[/tex]

[tex]\Longrightarrow\sf{\sqrt{x^{2}}=|x|\ (-x < 0)}[/tex]

[tex]\;[/tex]

So, we found out that the negative values also follow the result.

[tex]\boxed{\sf\therefore{\sqrt{(x)^{2}}=|x|\textsf{ for all }x\in\mathbb{R}}}[/tex]

[tex]\;[/tex]

Keep learning!

I'll give an answer from an analytic geometry point of view.

Explanation:

In analytic geometry, we define the distance of point P(a, b) from the point Q(c, d) using distance formula as,

[tex]\boxed{Distance = \sqrt{(c-a)^2 + (d-b)^2}}[/tex]

(Not giving the proof as it is well known. Can be easily proved using Pythagoras theorem.)

Considering P(0, 0) and Q(x, 0) gives the distance as,

[tex]Distance = \sqrt{(x-0)^2 + (0-0)^2} = \sqrt{x^2}[/tex]

From the definition of distance, we have that:

• Distance between two elements is actually defined as the measure of "absolute" length between them.

Since the distance between two points can't be negative, we can conclude that [tex]\sqrt{x^2} = |x|[/tex].

This can also be proved another way:

In complex analysis and vector analysis, we define [tex]|x|[/tex] as the distance of the complex number [tex]x[/tex] from origin and the magnitude of the position vector [tex]\vec{x}[/tex]respectively.

Again, from the definition of magnitude, we have that:

• Magnitude of any number [tex]x[/tex] is usually called its "absolute value" or modulus, denoted by [tex]|x|[/tex].

So, it would be fair to conclude that [tex]\boxed{\sqrt{x^2} = |x|\ \forall \ x\in \mathbb{R}}[/tex]..