floor function equality ⌊x⌋^2=16

Question

​

$$⌊x⌋^2=16 $$ $$⌊x⌋^2≤x^2≤⌊x⌋^2+1 $$ $$16≤x^2≤17 $$ $$-17<x<17 $$ not sure about the last step,is it correct? if so is it full answer? ty.

Answer 1

We have:

$\lfloor x\rfloor^{2} = 16$

$\lfloor x \rfloor = \pm 4$

Because the floor function outputs the greatest integer below or at the argument:

$\boxed{-4\leq x<-3\text{ or }4\leq x < 5}$

Answer 2

None of the steps are correct.

$[x]^2 \ne [x^2]$. Consider $x = 4.99$ then $[x]^2 = 4^2 = 16$ but $[x^2]=[4.99^2] = [24.9001] = 24$.

$16 \le x^2 < 17$ means $-17 \le x < 17$ is .... I don't get it.

if $16\le x^2 < 17$ were valid (which it isn't) the proper way to solve is $4 \le \sqrt{x^2} < \sqrt {17}$ so $4 \le |x| < \sqrt{17}$ and $-\sqrt{17} < x \le -4$ or $4 \le x < \sqrt{17}$. But $16 \le x^2 < 17$ is not correct.

Proper step is to simply solve $[x]$: If $[x]^2 = 16$ then $[x] =\pm 4$ and then simply continue as in Joshua Wang's answer: If $[x]=-4$ then $-4\le x < -3$ and if $[x]=4$ then $4\le x < 5$.