Find the interval on which $ x^{2} - \lfloor x \rfloor - 3 < 0 $ holds.

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $-1\le x<\sqrt{5}$.