$\lfloor x^2 +2x +7/4\rfloor =k ,\lfloor x^2 +x +7/2\rfloor=n$

Question

$$\lfloor x^2 +2x +7/4\rfloor +\lfloor x^2 +x +7/2\rfloor=3$$

find the set x

[x]=floor function .

my try :

$\lfloor x^2 +2x +7/4\rfloor =k ,\lfloor x^2 +x +7/2\rfloor=n$

$$k≤x^2+2x +7/4

$$n≤x^2+x +7/2

thank you very much!

Answer 1

$$(x+1)^2+\frac{3}{4}=x^2+2x+\frac{7}{4}$$

And

$$(x+\frac{1}{2})^2+\frac{13}{4}=x^2+x+\frac{7}{2}$$

So, $\text{min}{\text{floor}(x^2+x+\frac{7}{2})}=\text{floor}(\frac{13}{4})=3$.

And it follows we must have,

$$0 \leq (x+1)^2+\frac{3}{4}<1$$

And,

$$3 \leq (x+\frac{1}{2})^2+\frac{13}{4}<4$$

Both satisfied

Only concerning ourselves with real numbers this is the same as,

$$0 \leq (x+1)^2 < \frac{1}{4}$$

$$0 \leq (x+\frac{1}{2})^2 < \frac{3}{4}$$

Or,

$$|x+1| <\frac{1}{2}$$

$$|x+\frac{1}{2}|< \frac{\sqrt{3}}{2}$$

Which means we must satisfy,

$$-\frac{3}{2}< x< -\frac{1}{2}$$

$$-\frac{\sqrt{3}}{2}-\frac{1}{2}

Both are satisfied when we have,

$$-\frac{\sqrt{3}}{2}-\frac{1}{2}< x< -\frac{1}{2}$$