Find the set of all rea numbers x so that \(\lfloor x^2 \rfloor = 3\) Express your answer in interval notation
Floor basically means... rounding down to the nearest number.
Thus, we should set \(3 \leq x^2 < 4\).
Similarly, we can also set \(-3 \leq x^2 < -4\).
Taking the square root gives... \(\sqrt{3} \leq x < \sqrt{4}\) and \(-\sqrt{3} \leq x < -\sqrt{4}\).
Thus entering in integer notation \((-\sqrt{4}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{4})\). (square bracket = include end, round bracket = don't include end, U = union (include both in solution).