If y>0, find the range of all possible values of y such that y*floor(y) = 42. Express your answer using interval notation.
+9519 Suppose that \(y = r + x\), where \(0 \leq x < 1\) and r is an integer.
Then \(\lfloor y \rfloor = r\).
\(r(r +x) = 42\\ x = \dfrac{42}r - r\)
Then x has to satisfy the inequality \(0 \leq x < 1\).
We find the range of values of r such that \(0 \leq \dfrac{42}r - r < 1\).
Since y > 0, r >= 0. We multiply each term by r to get \(0 \leq 42 - r^2 < r\).
Solving \(0 \leq 42 - r^2\) gives \(-\sqrt{42} \leq r \leq \sqrt{42}\).
Solving \(42 - r^2 < r\) gives \(r > 6\text{ or }r < -7\).
Combining the solutions of the two inequalities gives \(6 < r \leq \sqrt{42}\). But this inequality has no integer solutions.
Therefore, there are no such value of y such that y * floor(y) = 42.
