quadratic equation with floor function.

[jagdish]

Find all Real solution that satisfy the quadratic equation $6x^2+77\lfloor x \rfloor +147=0$

Where $\lfloor x \rfloor =$ Floor function.

like $\lfloor 2.3 \rfloor =2$ OR $\lfloor -2.3 \rfloor =-3$

[jagdish]

any idea about that question

Thanks.

[Avik Roy]

I guess no rational solution exists.

Let,
$x = \lfloor {x} \rfloor + \frac {a} {b}$ where $a < b$ and $a,b$ are mutually coprime.
Plugging this in the given equation, we obtain-
$6 \left( \lfloor {x} \rfloor ^2 + 2\lfloor {x} \rfloor \frac {a}{b} + \frac {a^2}{b^2}\right) + 77 \lfloor {x} \rfloor +147 = 0$
For the equation to satisfy $6 \left(2\lfloor {x} \rfloor \frac {a}{b} + \frac {a^2}{b^2}\right) = \frac {6a \left( 2b\lfloor {x} \rfloor + a \right)}{b^2}$ must be an integer.
Since $a$ is coprime with $b$, any factor dividing $b$ can divide neither $2b\lfloor {x} \rfloor + a$ nor $a$. Hence, $b^2$ must divide $6$, that gives us only possible value $b = 1$ reducing $\frac {a}{b} = 0$.
Then we have to solve the equation $6x^2 + 77x + 147 = 0$ for integral $x$, which is not possible.

But I still can't go through the irrational wall

[nayel]

Let $\lfloor x\rfloor=n$ and $x=n+a$, with $0\le a<1$. Then our equation becomes $6(n+a)^2+77n+147=0$, i.e.
\[6a^2+12na+(6n^2+77n+147)=0.\]
This is a quadratic in $a$ with discriminant $144n^2-24(6n^2+77n+147)=-24(77n+147)$, which must be non-negative. Hence $n\le -147/77=-21/11$ i.e. $n\le -2$. On the other hand,
\[a=-n\pm\sqrt{\frac{-77n-147}{6}}.\]
If $\displaystyle a=-n+\sqrt{\frac{-154n-294}{12}}$ then $a\ge 2$, impossible. Hence we must have
\[a=-n-\sqrt{\frac{-77n-147}{6}}.\qquad(*)\]
Hence $0\le a<1$ implies
\[\begin{align*}&n\le -\sqrt{\frac{-77n-147}{6}}<n+1\\
\Rightarrow &n^2\ge\frac{-77n-147}{6}>(n+1)^2\\
\Rightarrow &6n^2+77n+147\ge 0>6n^2+89n+153
\end{align*}\]
Thus
\[-2\ge n\ge-\frac 73\quad\text{or}\quad n\le-\frac{21}{2}\Rightarrow n=-2\quad\text{or}\quad n\le-11\]
and
\[\frac{-89-\sqrt{89^2-24\cdot 153}}{12}<n<\frac{-89+\sqrt{89^2-24\cdot 153}}{12}\quad\text{i.e.}\quad -12\le n\le -2.\]
Therefore the possible values of $n$ are $-2,-11$ and $-12$, and for each of them we get from $(*)$, $\displaystyle a=\frac{12-\sqrt{42}}{6}, \frac{33-5\sqrt{42}}{3}, \frac{24-\sqrt{518}}{2}$. Thus
\[x=-2+\frac{12-\sqrt{42}}{6},\; -11+\frac{33-5\sqrt{42}}{3},\; -12+\frac{24-\sqrt{518}}{2}\]
i.e.
\[x=-\sqrt{\frac{7}{6}},\; -5\sqrt{\frac{14}{3}},\; -\sqrt{\frac{259}{2}}.\]

Hope I didn't make any obviuos mistakes throughout this mess...

[Avik Roy]

A big round of applause for Nayel for this awesome solution...commendable approach

[jagdish]

Thanks nayel for brilliant solution.