The solutions of the inequality $[x^2]+5[x]+6>2$?

Question

I tried to solve this inequality by taking the square outside the floor function $[y]$ (greatest integer less than $y$)but it was wrong since if $x=2.5$ then $[x]= 2$ and $x^2=4$ while $[x^2]=[6.25]=6$.

Answer 1

We can you Calvin's suggested method here (see his comment) by solving the equation as we would with the quadratic equation. You solve this equation as following, let us take $[x]$ another variable say $n$, and $[x^2]$ as $n^2 + c$. Now your equation becomes :

$$n^2 + c + 5n + 6 = 2$$ $$n (n + 5) = -4 - c$$

Now, we know from above equation that, n varies from , $$ n = -4 -c$$ to $$ n = -9 - c$$

here, for our required equation to be true, the equation should be strictly greater than $2$. Hence, the least possible value for $x$ in greatest integer function should be greater than $-4$.

I hope this clears your doubt.

Answer 2

Define $f(x)=\operatorname{floor}\left(x^2\right)+5\operatorname{floor}\left(x\right)+6 = [x^2] + 5[x] + 6$.

I'll present some graphs that indicate the solution set is more complicated than $x\ge -4$.

(Desmos link)

We see the solution is made up of 3 different regions. As you can see Desmos is not certain about the accuracy of its plot so I found a second opinion that indeed indicates that there is a small hole in the solution regions a small bit to the left of $-5$ (W|A link),

I don't understand where $-4$ came from. Firstly, it does not come from the other potential interpretation of $[\cdot]$ as the ceiling function, as these calculations in W|A show. Secondly, under the interpretation $[x] = \operatorname{floor}(x)$, we have $f(-2)=0$.

Clearly, even without a graph, there are at least two distinct regions that satisfy the inequality, since if $x>0$, then

$$ f\left(x\right)=[x^2] + 5[x]+6 > 6 > 2$$ and if $x < -6 $ then $$ [x^2] + 5[x]+6 ≥ (x^2-1) + 5(x-1) + 6 = \underbrace{x(x+5)}_{>6} > 2 $$ so the solution region must contain $$ \{ x < -6 \} \cup \{x > 0\}$$ as a subset. The fact that there is a third region is not so obvious and requires more careful study of the difference between $[x^2]$ and $x^2$.

Answer 3

I think the answer should be: $x\le-\sqrt{27}\qquad $ or $\qquad -5\le{x}\le-\sqrt{22}\qquad $ or $\qquad 0\le{x}\qquad $ Here is a sketch:

For $$-4\le{x}\le-1$$ it is $$x^2+5x+4\le0$$ or $$x^2+5x\le-4$$, but also $$ [x^2]+5[x]\le{x^2+5x}$$ so there is no solution in $[-4, -1]$.

For $$-5\le{x}\lt-4$$ it is $\qquad [x]=-5\qquad $ and so $\qquad 5[x]=-25\qquad $ therefor $\qquad [x^2]-25\gt-4\qquad $ $\iff$ $\qquad [x^2]\geqslant22\qquad $ $\iff$ $\qquad x^2\geqslant22\qquad $ $\implies$ $\qquad x\le{-\sqrt{22}}\qquad $