sum of squares of the roots of equation

Question

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: $$(q+r)^2-7q+5=0.$$

Answer 1

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 $$2 + 23 + 30 + 37 = 92.$$

Answer 2

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 $2,23,30,37$