Hjdcti

How to solve inequalities in which we have quadratic terms and greatest integer function.

$$ 2< x^2 -[x]<5$$.
[.] is greatest integer function.
Do we need to break into the cases as [0,1), [1,2) and so on?

Hint: use $$x-1<[x]leq x$$ and then solve some quadratic inequalites like

$$ x^2-5<[x] implies x^2-5<ximplies x^2-x-6<0$$

so $xin (-2,3)$ and so on...

• If $xin (-2,-1)$ then $[x]=-2 $ so $0<x^2<3$ so $-sqrt3<x<sqrt3$ so $xin(-sqrt3,-1)$
  


• If $xin [-1,0)$ ...

While breaking up into integer intervals always works, sometimes it is too much work.

Let's say you instead had $$
213 < x^2 - lfloor x rfloor < 505 $$
You really don't want to consider $24$ cases individually.

So you get clever and change this to a pair of simultaneous inequalities: Let $x = n + y$ with $n in Bbb Z$ and $0 leq y < 1$. Then the inequalities are
$$
left{ beginarrayl 213 < (n+y)^2 - n < 505 \ 0 leq y < 1 endarray right.
$$
Then $$213 < n^2 + (2y-1)n +y^2 , 0 leq y < 1 implies 213 < n^2 + (1)n + 1
$$
and this shows that $n geq 15$ (you can get that with one application of the quadratic formula), cutting out a lot of the work. Similarly,
$$ n^2 + (2y-1)n +y^2 < 505 , 0 leq y < 1 implies n^2 + (-1)n + 0 < 505
$$
and this shows that $n < 23$.

Finally, you can also justify only examining the allowed values of $y$ in the boundary cases ($n = 15$ and $n = 22$); in between, any value of $y$ works. (This might not be the case for cubic expressions, for example.)

So for example, on the low end, you would need to solve for $y$ in
$$
213 < (15+y)^2 - 15 = 210 - 30 y + y^2 \
y^2 -30 y -3 > 0\
y > frac12( sqrt912 -30 ) \
x > (sqrt228 - 15) + 15 implies x > sqrt228
$$
and similarly you need to consider the case of $n=22$ to get the upper limit for $x$.