floor function and check if my answer is correct

Question

$$ ⌊x^2⌋ =3 $$ $$⌊x^2⌋≤x^2<⌊x^2⌋+1 $$ $$ 3≤x^2<4 $$ $$ x^2<4 = \sqrt{4}<x<\sqrt{4} $$ $$ 3≤x^2 $$ $$ x\le \:-\sqrt{3}\quad \mathrm{or}\quad \:x\ge \sqrt{3} $$

Answer 1

It looks good, though i think you forgot a negative in you expansion of $x^2<4$,

as $x^2<4$ $⟺$ $-\sqrt{4}<x<\sqrt{4} $

which simplifies to $-2<x<2$.

so your solutions are all $x$ that satisfy that condition and the one on your last line, $ x\le \:-\sqrt{3}\quad \mathrm{or}\quad \:x\ge \sqrt{3}$

These conditions can be further tidied to, $-2<x≤-\sqrt{3}$ $ $ or $\sqrt{3}≤x<2$

but thats all notation, your solution looks to be correct

hopefully that helps

Answer 2

After you obtain $$3 \le x^2<4$$

We have $$3 \le |x|^2 < 4$$

$$\sqrt3 \le |x| < 2$$

Hence the solution is $-2 < x \le -\sqrt3$ or $ \sqrt3 \le x < 2$.

Comment about your approach:

Seems to have typos and after you obtain $-2 < x < 2$ and $(x \le -\sqrt3) \lor (x \ge \sqrt3)$, you might like to sketch the corresponding region on the number line to help you identify the region.