Problem
Solve the equation
$\frac{x^2 – 2x + 3}{\lbrace x\rbrace} + \frac{2023}{2}{\lbrace x\rbrace} = 2\lfloor x \rfloor + 88$,
where $\lfloor x \rfloor$ and $\{x\}$ are the greatest integer less than or equal to $x$ and the fractional part of $x$, respectively.
- This problem is from the Team Round of a local high school math competition that has ended recently
My work so far
$x$ can not be an integer because $\lbrace x\rbrace = 0$ will be dividing by zero.
Therefore, $x$ is not an integer.
We can express $x$ as $\lfloor x \rfloor + \lbrace x\rbrace$. Lets say $\lfloor x \rfloor = a$, and $\lbrace x\rbrace = b$.
So $x = a + b.$
We have:
$((a+b)^2 – 2(a+b) + 3) + \frac{2023}{2} b^2 = 2ab + 88b$
which turns into:
$2a^2 + 2025b^2 – 4a -180b + 6 = 0$
(Hopefully I did my math correctly)
I am not too sure what to do from here on. Am I going in the right direction? Am I supposed to solve this equation? Thank you for your help.
Best Answer
You were on the right track. Take your last equation and complete the square
$$2\Bigr(\lfloor x\rfloor-1\Bigr)^2 + \Bigr(45\{x\}-2\Bigr)^2 = 0$$
The only way the sum of two nonnegative terms is $0$ is if they were both $0$ individually. This leaves us with the answer
$$\begin{cases}\lfloor x\rfloor = 1 \\ \{x\} = \frac{2}{45}\end{cases} \implies \boxed{x = \frac{47}{45}}$$