Floored Vieta’s

Rewrite the equation as

$\lfloor x \rfloor = \dfrac{4x^2 + 51}{40}.$

Let $t = \lfloor x \rfloor.$ By the equality above, we can write $x$ as

$x = \dfrac{\sqrt{40t - 51}}{2}.$

Note that we take the positive square root because if $x < 0,$ then $4x^2 - 40 \lfloor x \rfloor + 51 > 0.$

To find bounds on $t,$ we use the fact that $\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1.$ Substituting our value for $t$ into this inequality, we get

$\begin{aligned} t \leq \dfrac{\sqrt{40t - 51}}{2} \\ 4t^2 - 40t + 51 \leq 0 \\ (2t - 3)(2t - 17) \leq 0 \\ t \geq \dfrac{3}{2} \quad \text{and} \quad t \leq \dfrac{17}{2} \end{aligned}$

and

$\begin{aligned} \dfrac{\sqrt{40t - 51}}{2} < t + 1 \\ 4t^2 - 32t + 55 > 0 \\ (2t - 11)(2t - 5) > 0 \\ t < \dfrac{5}{2} \quad \text{or} \quad t > \dfrac{11}{2}. \end{aligned}$

The combined bounds for $t$ are $t \in \left [\frac{3}{2}, \frac{5}{2} \right ) \cup \left (\frac{11}{2}, \frac{17}{2} \right ].$ Since $t$ is the floor of a number, it is always an integer, so the possible values of $t$ are 2, 6, 7, and 8.

Plugging these values of $t$ back into our original expression for $x$ yields these four solutions:

$\begin{aligned} x &= \dfrac{\sqrt{40(2) - 51}}{2} = \dfrac{\sqrt{29}}{2} \\ x &= \dfrac{\sqrt{40(6) - 51}}{2} = \dfrac{3 \sqrt{21}}{2} \\ x &= \dfrac{\sqrt{40(7) - 51}}{2} = \dfrac{\sqrt{229}}{2} \\ x &= \dfrac{\sqrt{40(8) - 51}}{2} = \dfrac{\sqrt{269}}{2}. \end{aligned}$

The sum of the squares of these solutions is

$\left ( \dfrac{\sqrt{29}}{2} \right )^2 + \left ( \dfrac{3 \sqrt{21}}{2} \right )^2 + \left ( \dfrac{\sqrt{229}}{2} \right )^2 + \left ( \dfrac{\sqrt{269}}{2} \right )^2 = \dfrac{29 + 189 + 229 + 269}{4} = \boxed{179}.$

$\begin{aligned} 4x^2 - 40\lfloor x \rfloor + 51 & = 0 \end{aligned}$

$\begin{aligned} x & = \sqrt{10\lfloor x \rfloor - 12.75} & \small \color{#3D99F6} \text{Note that }x = \lfloor x \rfloor + \{x\} \\ \implies \{x\} & = \sqrt{10\lfloor x \rfloor - 12.75} - \lfloor x \rfloor & \small \color{#3D99F6} \text{Note that } 0 \le \{x\} < 1 \end{aligned}$

$\begin{aligned} \implies 0 \le \sqrt{10\lfloor x \rfloor - 12.75} & - \lfloor x \rfloor < 1 \\ \lfloor x \rfloor \le \sqrt{10\lfloor x \rfloor - 12.75} & < \lfloor x \rfloor + 1 \\ \lfloor x \rfloor^2 \le 10\lfloor x \rfloor - 12.75 & < \lfloor x \rfloor^2 +2 \lfloor x \rfloor + 1 \end{aligned}$

$\implies \begin{cases} \lfloor x \rfloor^2 - 10\lfloor x \rfloor + 12.75 & \le 0 & \implies (\lfloor x \rfloor -1.5)(\lfloor x \rfloor -8.5) \le 0 & \implies 2 \le \lfloor x \rfloor \le 8 \\ \lfloor x \rfloor^2 - 8\lfloor x \rfloor + 13.75 & > 0 & \implies (\lfloor x \rfloor -2.5)(\lfloor x \rfloor -5.5) > 0 & \implies \lfloor x \rfloor \le 2 \cup \lfloor x \rfloor \ge 6 \end{cases}$

$\implies \lfloor x \rfloor = \begin{cases} 2 & \implies x^2 = 10(2) - 12.75 = 7.25 \\ 6 & \implies x^2 = 10(6) - 12.75 = 47.25 \\ 7 & \implies x^2 = 10(7) - 12.75 = 57.25 \\ 8 & \implies x^2 = 10(8) - 12.75 = 67.25 \end{cases}$

The sum of the squares of all real solutions is $7.25+47.25+57.25+67.25 = \boxed{179}$ .

We know $[x]\leq x$

Note that $[x]$ cannot be negative.

$\Rightarrow -40[x]\geq -40x$

$\Rightarrow 4x^{2}-40[x]+51\geq 4x^{2}-40x+51$

Let $f(x)=4x^{2}-40[x]+51$

and $g(x)=4x^{2}-40x+51$

$\Rightarrow f(x)\geq g(x)$

For $x\geq 10$

$\Rightarrow g(x)\geq 51$

$\Rightarrow f(x)\geq 51$ as $f(x)\geq g(x)$

But we want the value of $x$ such that $f(x)=0$ which is less than 51

$\Rightarrow x<10$

$\Rightarrow [x] \in {0,1,2,3,4,5,6,7,8,9}$

Since $4x^{2}=40[x]-51$

Hence $x^{2} \in {\frac{-51}{4},\frac{-11}{4},\frac{29}{4},\frac{69}{4},\frac{109}{4},\frac{149}{4},\frac{189}{4},\frac{229}{4}\frac{269}{4},\frac{309}{4}}$

Since $x^{2}\geq0$ Therefore $x^{2} \in \frac{29}{4},\frac{69}{4},\frac{109}{4},\frac{149}{4},\frac{189}{4},\frac{229}{4}\frac{269}{4},\frac{309}{4}$

So now we have $x\in \sqrt[ ]{\frac{29}{4}}\sqrt[ ]{\frac{69}{4}}, \sqrt[ ]{\frac{109}{4}},\sqrt[ ]{\frac{149}{4}},\sqrt[ ]{\frac{189}{4}},\sqrt[ ]{\frac{229}{4}},\sqrt[ ]{\frac{269}{4}},\sqrt[ ]{\frac{309}{4}}$

Checking against the equation leaves $x\in \frac{\sqrt{29}}{2}, \frac{\sqrt{189}}{2},\frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}$

Therefore the sum of squares is $\frac{29}{4}+\frac{189}{4}+\frac{229}{4}+\frac{269}{4}=\boxed{179}$