6. [5] What is the sum of the positive solutions to $2 x^{2}-x\lfloor x\rfloor=5$, where $\lfloor x\rfloor$ is the largest integer less than or equal to $x$ ?
Answer: $\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$ We first note that $\lfloor x\rfloor \leq x$, so $2 x^{2}-x\lfloor x\rfloor \geq 2 x^{2}-x^{2}=x^{2}$. Since this function is increasing on the positive reals, all solutions must be at most $\sqrt{5}$. This gives us 3 possible values of $\lfloor x\rfloor: 0,1$, and 2 .
If $\lfloor x\rfloor=0$, then our equation becomes $2 x^{2}=5$, which has positive solution $x=\sqrt{\frac{5}{2}}$. This number is greater than 1 , so its floor is not 0 ; thus, there are no solutions in this case.
If $\lfloor x\rfloor=1$, then our equation becomes $2 x^{2}-x=5$. Using the quadratic formula, we find the positive solution $x=\frac{1+\sqrt{41}}{4}$. Since $3<\sqrt{41}<7$, this number is between 1 and 2 , so it satisfies the equation.
If $\lfloor x\rfloor=2$, then our equation becomes $2 x^{2}-2 x=5$. We find the positive solution $x=\frac{1+\sqrt{11}}{2}$. Since $3<\sqrt{11}<5$, this number is between 2 and 3 , so it satisfies the equation.
We then find that the sum of positive solutions is $\frac{1+\sqrt{41}}{4}+\frac{1+\sqrt{11}}{2}=\frac{3+\sqrt{41}+2 \sqrt{11}}{4}$.