sum of squares of the roots of equation The equation is $$x^2-7[x]+5=0.$$
Here $[x]$ the greatest integer less than or equal to $x$. Some other method other than brute forcing. I tried a method of putting $[x]=q$ and $x=q+r$ which gives an equation:  $$(q+r)^2-7q+5=0.$$
 A: Set $[x] = q$ and solve for $x$:
$$ x = \pm \sqrt{7 q - 5}.$$
So $q \geq 1$.  Now use that $0 \leq x - q < 1$ to conclude that $x$ must be positive and $$0 \leq \sqrt{7 q - 5} - q < 1.$$ This restricts $q$ to $q \in \{1,4,5,6\}$ and the roots are therefore $\sqrt{2}, \sqrt{23}, \sqrt{30}, \sqrt{37}$.  The sum of the squares of the roots is therefore $$2 + 23 + 30 + 37 = 92.$$
A: HINT:
$0=x^2-7[x]+5\ge x^2-7x+5$
So, $x^2-7x+5\le0$
If $a,b(>a)$ are the roots of $x^2-7x+5=0, a\le x\le b$
Now, $a,b$ are $\approx .8,6.19$
Now, testing for the cases
As $x^2=7[x]-5$
If $0<x<1,[x]=0\implies x^2+5=0$ which does not have any real solution 
If $1<x<2,[x]=1\implies x^2=7-5=2$
If $2<x<3,[x]=2\implies x^2=14-5=9$ but $x$ can not be integer greater than $[x]$
If $3<x<4,[x]=3\implies x^2=7\cdot3-5=16$ but $x$ can not be integer greater than $[x]$
If $4<x<5,[x]=4\implies x^2=7\cdot4-5=23$
If $5<x<6,[x]=5\implies x^2=7\cdot5-5=30$
If $6<x<7,[x]=6\implies x^2=7\cdot6-5=37$
If $7<x<8,[x]=7\implies x^2=7\cdot7-5=44<7^2$
So, the values of $x^2$ are $2,23,30,37$
