Why does $(x-2)^2\geq 1$ imply that $|x-2|\geq\sqrt1$ or $|x-2|\leq -\sqrt1$

inequality

I want to improve my theoretical background to make it rock solid.

Given

$(x – 2)^2 \geq 1$

O learnt that taking the square root of boot sides yields

$| x -2| \geq \sqrt{1}$ or $| x – 2 | \leq – \sqrt{1}$

Here, I have two questions:

  1. Why does taking the square root transforms it from $(something)^2$ to $|something|$? Why the "module" symbols?

    I believe this is redundant, not really necessary. Am I right?

and

  1. Why does the sign change direction?

    I know this is because $x$ can be $ \le 0$, so I need to go into a little bit more depth about the theory on swapping signs. As far as I can tell we only swap inequality sign when dividing or multiplying by a negative number

Thanks a lot in advance!

Best Answer

The chain of equivalent transformations should progress first to $$ |x-2|\ge\sqrt1=1 $$ and then to $$ x-2\le -1\text{ or }x-2\ge 1. $$

If $|u|\ge a$, then

  • either $u\ge 0$ and $u=|u|\ge a$.
  • or $u<0$ and $-u=|u|\ge a\iff u\le -a$.

You can go from $u^2\ge a^2$, $a>0$, to $|u|\ge a$ because the square function is strictly monotonically increasing on the positive half-axis. Or just simply because in $$0\le u^2-a^2=(|u|-a)(|u|+a)$$ you can divide out the positive second factor.

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