Consider the following problem:
Find the radius of convergence of:
$$\sum_{n=0}^\infty \left({(x-4)^2\over 3}\right)^n$$
This is a geometric series, from which you can get the inequality:
$$\left|\left({(x-4)^2\over 3}\right)\right| < 1$$
From here, my teacher rewrote and solved the quadratic as follows:
$$-1 < {(x-4)^2\over 3} < 1 \implies -3 < {(x-4)^2} < 3$$
From there, it was split into two, and the former is always true: $$\require{cancel} \cancel{-3 < {(x-4)^2}} \quad {(x-4)^2} < 3 \implies x^2-8x+13 < 0$$ $$x = {8\pm\sqrt{64-4(1)(13)}\over 2} \implies x={4\pm\sqrt3}$$
Then, by testing intervals, the radius was found to be $\sqrt3$ and the interval $4-\sqrt{3} < x < 4+\sqrt3$. I thought this was a bit tedious, so I tried to find the answer without solving quadratics. My question is, is the following method algebraically valid, as this is a much faster way to solve the problem:
$$\left|\left({(x-4)^2\over 3}\right)\right| < 1 \implies \left|(x-4)^2\right| < 3$$
Since $3$ is positive, this should be valid. Further:
$$\left|(x-4)^2\right| < 3 \implies \left|x-4\right|^2 < 3$$
I'm not sure about this step since I don't know if it's rigorously correct, but it seems to be equivalent because they both result in always positive outputs so the order shouldn't matter right? Building on this:
$$\left|x-4\right|^2 < 3 \implies |x-4|<\sqrt3 \implies -\sqrt3 < x-4 < \sqrt3$$
Which leads the same radius and interval. Is this method legal?
It's much simpler than that, if you rember that $\;\sqrt{u^2}=|u|$: $$(x-4)^2<3\iff \sqrt{(x-4)^2}<\sqrt 3,\enspace\text{i.e. }\enspace |x-4|<\sqrt 3\enspace\text{ or }\enspace 4-\sqrt 3<x<4+\sqrt 3.$$