Find the interval on which $ x^{2} - \lfloor x \rfloor - 3 < 0 $ holds. On what interval does the equation $ x^{2} - \lfloor x \rfloor - 3 < 0 $ hold?
My attempt: I tried sketching the graph, but it’s a bit complicated.
Is there any other approach?
 A: Hint: We have $0\le\big\{x\big\}<-[x]+\sqrt{[x]+3}~,~$ with $[x]\in$ Z and $\big\{x\big\}<1$, if x is positive, and $1>\big\{x\big\}>-[x]-\sqrt{[x]+3}~,~$ with $[x]\in$ Z and $0\le\big\{x\big\}$, assuming x is negative.
A: Probably graphing $y=x^2-3$ and $y=\lfloor x\rfloor$ and seeing where the first graph is below the second is the easiest way to solve this, but here is another method:
$\textbf{1)}$ Since $\lfloor x\rfloor \le x$,
$x^2-3<\lfloor x\rfloor\implies x^2-3<x\implies x^2-x-3<0\implies x^2-x+\frac{1}{4}<\frac{13}{4}\implies$
$\left(x-\frac{1}{2}\right)^2<\frac{13}{4}\implies|x-\frac{1}{2}|<\frac{\sqrt{13}}{2}<2\implies-\frac{3}{2}<x<\frac{5}{2}$.
$\textbf{2)}$ Since $x-1<\lfloor x\rfloor$,  the inequality will be satisfied if $x^2-3\le x-1$:
$x^2-3\le x-1 \iff x^2-x-2\le0\iff (x-2)(x+1)\le 0\iff -1\le x\le2$.
$\textbf{3)}$ a) If $-\frac{3}{2}<x<-1$, then $\lfloor x\rfloor=-2$, 
so $x^2-3<\lfloor x\rfloor \iff x^2<1\iff -1<x<1$; so there is no solution in this case.
b) If $2<x<\frac{5}{2}$, then $\lfloor x\rfloor=2$, so $x^2-3<\lfloor x\rfloor \iff x^2<5\iff-\sqrt{5}<x<\sqrt{5}$;
and therefore if $2<x<\sqrt{5}$, the inequality is satisfied.
Therefore the inequality is satisfied for $-1\le x<\sqrt{5}$.
A: The graph is a curve.  You need the points where thr graph cuts the y-axis.  Solve for eq=0 and find the two values  for x=(1+sqrt(13))/2 and 
x=(1-sqrt(13))/2
used -b formula. 
this gives you your x-axis points where y=0. 
Now because the graph has a positive x squared coefficient. It is an n shaped curve.
now graph an n-shaped curve that goes through the two points above.
You should be able to work out when the graph is <0.  
It will be something like when 
x<(1-sqrt(13))/2 and x>(1+sqrt(13))/2. 
I could be wrong about the soln.  I need to write it out.  
