quadratic equation with floor function..

[man111]

find all real values of [tex]x[/tex] that satisfy the equation [tex]6x^2+77 \lfloor\ x \rfloor\[/tex] + 147 = 0.

Where [tex]\lfloor\ x \rfloor\ =[/tex]floor function.

like [tex]\lfloor\ 2.3 \rfloor\ = 2[/tex] and [tex]\lfloor\ -2.3 \rfloor\ = -3[/tex]

[martin123456]

[tex]6x^2+147=-77\lfloor{x}\rfloor[/tex]. The RHS is an integer, so LHS must also be an integer. That leads to [tex]6x^2[/tex] being an integer. So [tex]x^2=\frac{a}{6}[/tex] where [tex]a \in \mathbb{Z}[/tex].
1) [tex]x \in \mathbb{Z}[/tex]. Then the equation is [tex]6x^2+77x+147=0[/tex]. [tex]x_{1,2}=\frac{-77 \pm 49}{12}[/tex]. Check if some of the roots is an integer.
2) [tex]x \in \mathbb{Q}[/tex]. Let [tex]x=\frac{p}{q}[/tex], where [tex]p,q \in \mathbb{Z}[/tex] and [tex](p,q)=1[/tex]. We have that [tex]\frac{p^2}{q^2}=\frac{a}{6} \Leftrightarrow 6p^2=aq^2[/tex]. That means that [tex]q^2|6[/tex], so there are no options for [tex]q[/tex] (remember [tex]|q| \ne 1[/tex]).
3) [tex]x \in \mathbb{R}[/tex]. Then [tex]a+147+77\lfloor{\pm \sqrt{\frac{a}{6}}}\rfloor=0[/tex]. Since [tex]a > 0[/tex] we must have [tex]x=- \sqrt{\frac{a}{6}}[/tex]. So the equation is [tex]a+147-77 \lfloor{\sqrt{\frac{a}{6}}}\rfloor = 0[/tex].
Let [tex]\lfloor{\sqrt{\frac{a}{6}}}\rfloor = b \in \mathbb{N} \Rightarrow b \le \sqrt{\frac{a}{6}} < b+1 \Rightarrow 6b^2 \le a < 6(b+1)^2[/tex]. We have [tex]a+147=77b \Rightarrow a = 77b-147[/tex]. SO the following inequalities hold: [tex]6b^2 \le 77b-147 \le 6(b+1)^2[/tex]. The first ineq is [tex]\Leftrightarrow 6b^2-77b+147 \le 0 \Leftrightarrow D = 49[/tex] so the roots are [tex]b_{1,2}=\frac{77 \pm 49}{12}[/tex], [tex]b_1=\frac{7}{3}[/tex], [tex]b_2=\frac{31}{3}[/tex], so [tex]b \in \{3,4,\ldots,10\}[/tex]. From the other inequality we get that [tex]b[/tex] is either less than 4 or bigger than 7, so values for [tex]b[/tex] are 3,8,9,10.
Now it's very easy to find the solutions. For example, let [tex]b=3[/tex]. Then the equation is [tex]a+147-231=0 \Leftrightarrow a = 84[/tex]. Now check if [tex]\lfloor{\sqrt{\frac{84}{6}}}\rfloor = 3[/tex]. It is so we have found one solution. Check for the others.

[man111]

martin123456 thanks for nice approach.