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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 $[x^2]=[6.25]=6$.

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    $\begingroup$ write $n = [x]$ and $[x^2] = n^2 + c$. Solve the problem as if its a quadratic in $n$, with answer depending on $c$. This gives you the range of values that $x$ can take $\endgroup$ – Calvin Khor Sep 7 '18 at 7:44
  • $\begingroup$ I think there are negative values for $x$. $\endgroup$ – mrs Sep 7 '18 at 7:54
  • $\begingroup$ What operation is your $[\cdot]$? $\endgroup$ – mvw Sep 7 '18 at 8:01
  • $\begingroup$ Oviously the roots are negative but what are they? $\endgroup$ – priyanka kumari Sep 7 '18 at 8:19
  • $\begingroup$ Did you try Calvin's method? When you take $[x]$ as $n$ and $[x^2]$ as $n^2$ and solve it as quadratic equation you'll get $n$ varying from $(-4-c)$ to $(-9-c)$. So, for the equation to be true your $n$ needs to be at least greater than $-4$. $\endgroup$ – Iti Shree Sep 7 '18 at 9:36
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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.

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  • $\begingroup$ Please tell me about Calvin method $\endgroup$ – priyanka kumari Sep 7 '18 at 11:22
  • $\begingroup$ I used the same thing he mentioned in the comment. $\endgroup$ – Iti Shree Sep 7 '18 at 11:49
  • $\begingroup$ If you make a graph of the function there should be 3 regions. One is the one you found, one more is for X sufficiently large and negative, and one more comes from a careful study of the term $c$ $\endgroup$ – Calvin Khor Sep 7 '18 at 13:15
  • $\begingroup$ You can try to write $[x]+d = x $ so that $n^2 +2dn+d^2=x^2 $ ie $c=[2dn+d^2], where d is any number in $[0,1)$. Some case analysis will be needed... $\endgroup$ – Calvin Khor Sep 7 '18 at 13:37
  • $\begingroup$ Actually it doesn't seem that any of the boundaries of the regions is -4...sorry $\endgroup$ – Calvin Khor Sep 7 '18 at 13:40
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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)graph

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),

enter image description here

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 $x^2$.

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  • $\begingroup$ I think you are copied the wrong question. $\endgroup$ – priyanka kumari Sep 8 '18 at 10:19
  • $\begingroup$ @user584880 please explain? $\endgroup$ – Calvin Khor Sep 8 '18 at 10:20
  • $\begingroup$ It was$$[x^2]+5[x]+6>2$$ $\endgroup$ – priyanka kumari Sep 8 '18 at 10:22
  • $\begingroup$ @user584880 what did I use instead? Looks the same to me? $\endgroup$ – Calvin Khor Sep 8 '18 at 10:35
  • $\begingroup$ @user584880 I made a large number of claims. If you could point out one that isn't right, I would appreciate it greatly. In addition, I have provided a number of graphs, and you can change the inputs. If you could provide me with a graph / link that has the 'correct' function, that would be great. $\endgroup$ – Calvin Khor Sep 8 '18 at 14:55
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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 $

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  • $\begingroup$ How did you obtain this answer? $\endgroup$ – Atrey Desai Sep 8 '18 at 18:49
  • $\begingroup$ For $x$ non negative or smaller than $-6$ is rather obvious. Then I checked the values in the interval $[-6, 0]$ $\endgroup$ – dmtri Sep 8 '18 at 18:52
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    $\begingroup$ this matches my graph :) $\endgroup$ – Calvin Khor Sep 9 '18 at 3:36

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