Square root of a squared number changes sign, which to apply first?

Question

Heres something Ive always found interesting. Supose we have a variable $x$, and $x$ equals a negative number:

Say:

$$x=-17$$

Now, I can apply a square to both sides of the equation and preserve the equality:

$$x^2=(-17)^2$$

Now I can apply the square root (or a power of $1/2$) to both sides and preserve the equality:

$$√x^2=√(-17)^2$$

Now here comes the thing that confuses me. If I go the algebraic way, the square root cancells the square, and we are back to:

$$x=-17$$

However if I first apply the square to -17 :

$$√x^2=√289$$

$$x=±17$$

Or if I take only the positive value : $|x|=17$

So you get 2 solutions for x depending on the order of the steps taken

So my question is this, which step should be taken first?

Secondary related question:

Take for instance a complex function with no imaginary part, say:

$$F(w)=-\sin(wT)$$

To take the magnitude of $F(w)$ I have to take the absolute value, or the square root of the square of the real part plus the square of the imaginary part.

Since there is no imaginary part:

$$|F(w)|=√-\sin(wT)^2+0^2$$

So which one of this is the correct answer?

$$|F(w)|=-\sin(wT)\text{ or }|F(w)|=|\sin(wT)|$$

Answer 1

General convention is to take $\sqrt{x^2}=|x|$ but when solving for equations both values can be considered and the values only satisfying original can be selected, especially the case with inverse trigonometric equations when multiple angles arise due to squaring and such operations.

Answer 2

This is an answer to your first question. In general it holds that $$ \sqrt{x^2}=|x|,$$ so that $$x = -17\implies x^2=(-17)^2=17^2 \implies \sqrt{x^2} = |x| = 17,$$ which means that $$x=17 \hspace{5px}\text{ or }\hspace{5px}x=-17. $$ But this actually introduces a certain phenomenon. It turns out that the equation derived from squaring both sides is not necessarily completely the same as the original equation. This is due to the fact that squaring an equation can lead to extraneous solutions. In other words: the square root actually doesn't completely cancel out the square, since $$\sqrt{x^2} = |x| \hspace{5px}\text{ and not }\hspace{5px}\sqrt{x^2}=x $$

In this example it is seen that a new solution $x = 17$ has appeared and it is not a solution to the original equation. So all solutions must be checked in the original equation.

Answer 3

Indeed we have: $$x=-17\Rightarrow x^{2}=\left(-17\right)^{2}\Rightarrow\sqrt{x^{2}}=\sqrt{\left(-17\right)^{2}}$$ However it shortsighted to say that 'square root cancels square'. In fact we have: $\sqrt{x^{2}}=\left|x\right|$ and not in general $\sqrt{x^{2}}=x$. Applying that we arrive at: $$x=-17\Rightarrow x^{2}=\left(-17\right)^{2}\Rightarrow\left|x\right|=17$$ wich is correct.

The arrow in the second implication can be turned around because $\left|x\right|=17$ implies that $\left|x\right|^{2}=17^{2}$ and we have $x^{2}=\left|x\right|^{2}$ and $17^{2}=\left(-17\right)^{2}$.

However the arrow in the first implication cannot be turned around: if $x=17$ then indeed $x^{2}=\left(-17\right)^{2}$ but $x=-17$ is not true.