Why is the floor implemented?

We can subtract $6\{ x\}$ from both sides of the equation, where $\{x\}$ is the fractional part of $x$ . We get

$x^2 - 6x + 5 = - 6\{x\}$

We know $-6\{x\}$ has the range $[0, -6)$ , so we find the values of $x$ for which the LHS lies in this range.

$x^2 - 6x + 5$ can be factorized to $(x -1 )(x - 5)$ . Its minimum will occur at $x = \frac{1 + 5}{2} = 3$ . Its minimum value is $-4$ which is greater than $-6$ , so for $x \in [1, 5], (x-1)(x-5)$ will lie in $[0, -6)$ , and we will find roots only in this interval.

---

We can now rewrite the original equation as $x^2 = 6 \lfloor x \rfloor - 5$ .

For $x \in [1, 2)$ , $\lfloor x \rfloor = 1 \implies x^2 = 6 \times 1 - 5 \iff x = +1$ .

For $x \in [2, 3)$ , $\lfloor x \rfloor = 2 \implies x^2 = 6 \times 2 - 5 \iff x = +\sqrt{7}$

For $x \in [3, 4)$ , $\lfloor x \rfloor = 3 \implies x^2 = 6 \times 3 - 5 \iff x = +\sqrt{13}$

For $x \in [4, 5)$ , $\lfloor x \rfloor = 4 \implies x^2 = 6 \times 4 -5 \iff x = +\sqrt{19}$

For $x = 5$ , $\lfloor x \rfloor = 5 \implies x^2 = 6 \times 5- 5 \iff x = +5$ is a root.

(We have considered only the positive roots since in case $x$ lies in a positive interval.)

Sum of squares of roots is $1 + 7 + 13 + 19 + 25 = \boxed{65}$ . $~_\square$

Rewrite the equation as follows:

$x^2=6\left\lfloor x\right\rfloor-5$

Now, by the trivial inequality , we have,

$6\left\lfloor x\right\rfloor-5=x^2\geq 0\implies \left\lfloor x\right\rfloor\geq\frac 56=0.8\overline{3}$

But since the floor function only returns integer values, we should have,

$\left\lfloor x\right\rfloor\geq\left\lceil 0.8\overline{3}\right\rceil=1\iff x\geq 1$

Further, note that if $x\in [m,m+1)$ for some positive integer $m$ , we have,

$x^2\in [m^2,(m+1)^2)\quad\textrm{and}\quad 6\left\lfloor x\right\rfloor-5=6m-5$

For equality to hold between the two expressions, we must have,

$m^2\leq 6m-5\lt (m+1)^2\iff 1\leq m\leq 5$

The above mentioned inequality is obtained by solving the inequality with the lower bound as follows:

$m^2\leq 6m-5\iff (m-5)(m-1)\leq 0$

The rest follows by using the fact that $ab\leq 0$ implies that one of $a,b$ is $\leq 0$ and the other is $\geq 0$ . You get two cases. Reject the absurd case and give the solution set as $m\in [1,5]$ using the correct case.

The upper bound is always maintained regardless of the value of $m$ because $(m+1)^2-6m+5=(m-2)^2+2\geq 2\gt 0~\forall~m\in\Bbb{Z^+}$ by the trivial inequality.

Hence, from all this, we can conclude that we have, for solutions $x$ to this equation,

$x^2=6m-5~\forall~1\leq m\leq 5~\land~m\in\Bbb{Z^+}$

The required sum can be evaluated using the result above as,

$\begin{aligned}\sum_x x^2=\sum_{m=1}^5(6m-5)&=6\left(\sum_{m=1}^5m\right)-5\times 5\\&=6\cdot\frac{5\times 6}{2}-25=90-25=\boxed{65}\end{aligned}$