What is the sum of the squares of the roots of the equation $x^2 − 7[x] + 5 = 0?$ (Here $[x]$ denotes the greatest integer less than or equal to $x$) I tried many different approaches to this question,
$α^2 = 7[α] - 5$ and 
$β^2 = 7[β] - 5$
So combining both the equations we get,
$α^2 + β^2 = 7([α] + [β]) - 10$,
Which is the answer. However I am unable to simplify the LHS.
 A: Note that there are more than two roots (the equation is not a quadratic polynomial). We solve the problem in
three steps.
1) For $x<1$, $x^2+5>0\geq 7[x]$. No roots in $(-\infty,1)$.
2) For $x\geq 7$, $x^2+5>7x\geq 7[x]$. No roots in $[7,+\infty)$.
3) Try to solve the equation for $x\in [1,7)$ where $[x]\in\{1,2,3,4,5,6\}$.
If $x\in[1,2)$ then $x^2=7[x]-5=2$ implies $x=\sqrt{2}\in [1,2)$. So $x=\sqrt{2}$ is a root.
If $x\in[2,3)$ then $x^2=7[x]-5=9$ implies $x=\sqrt{3}\not\in [2,3)$. No roots.
If $x\in[3,4)$ then $x^2=7[x]-5=16$ implies $x=4\not\in [3,4)$. No roots.
If $x\in[4,5)$ then $x^2=7[x]-5=23$ implies $x=\sqrt{23}\in [4,5)$.  So $x=\sqrt{23}$ is a root.
If $x\in[5,6)$ then $x^2=7[x]-5=30$ implies $x=\sqrt{30}\in [5,6)$.  So $x=\sqrt{30}$ is a root.
If $x\in[6,7)$ then $x^2=7[x]-5=37$ implies $x=\sqrt{37}\in [6,7)$.  So $x=\sqrt{37}$ is a root.
Finally the sum of the squares of the roots is $2+23+30+37=92$.
A: $x^2–7[x]+5=0$
$\Rightarrow x^2=7[x]–5 \hspace{1cm}(1)$
$\Rightarrow x^2≤7x–5$ [Since $[x]≤x$]
$\Rightarrow x^2–7x+5≤0$
Therefore, $0.8≤x≤6.2$
i.e., $0≤[x]≤6$
Since for $[x]=0$, equation $(1)$ becomes, $x²=–5$, which is not possible.
Therefore, $1≤[x]≤6$.
In equation $(1)$
For $[x]=1, x^2=2 \Rightarrow x=\pm\sqrt{2}$
For $[x]=2, x²=9 \Rightarrow x=\pm\sqrt{3}$ [not possible because $x=\pm3$ and $[x]=2$ doesn’t make any sense]
For $[x]=3, x²=16 \Rightarrow x=\pm\sqrt{4}$ [not possible because $x=±4$ and $[x]=3$ doesn’t make any sense]
For $[x]=4, x²=23 \Rightarrow x=\pm\sqrt{23}$
For $[x]=5, x²=30 \Rightarrow x=\pm\sqrt{30}$
For $[x]=6, x²=37 \Rightarrow x=\pm\sqrt{37}$
Therefore, sum of all the roots of the given equation $=2+23+30+37=92$.
Hope it helps!!!!
A: Let $\lfloor x\rfloor=:k$. Then we want an overview over the set of all pairs $(x,k)$ satisfying
$$x^2-7k+5=0, \quad x-1<k\leq x,\quad k\in{\mathbb Z}\ .$$
To this end set up an $(x,k)$-plane, draw the parabola $k={1\over7}(x^2+5)$, as well as the strip $x-1<k\leq x$. Now stare at this figure.
A: Assume $x = m+f$ when $m = [x]$ and $f = \{x\}$. Substitute the values in original equation to get $$f^2 +(2m)f + (m^2 -7m+5) = 0$$ Solving the quadratic we get $$\boxed{f = -m±\sqrt{7m-5}}$$ $m$ can take integral values starting from $1$ and $f$ is according to this inequality: $0≤f<1$. Fulfilling these conditions we get $(m,f) ≡ (1, \sqrt{2} -1),(4,\sqrt{23}-4),(5, \sqrt{30}-5),(6,\sqrt{37}-6)$. Therefore required answer is $2+23+30+37 = 92$.
