53. How many integers in the inequality vert x^2-xvert gt 2x Does it not apply? E) 6 A) 2 B) 3 C) 4 (D)

To solve the inequality \(|x^2 - x| > 2x\), we need to consider two cases based on the absolute value expression.<br /><br />**Case 1:** \(x^2 - x > 2x\)<br /><br />This simplifies to:<br />\[ x^2 - 3x > 0 \]<br />Factor the quadratic:<br />\[ x(x - 3) > 0 \]<br /><br />The critical points are \(x = 0\) and \(x = 3\). We test intervals around these points:<br /><br />- For \(x < 0\), choose \(x = -1\): \((-1)(-1 - 3) = 4 > 0\)<br />- For \(0 < x < 3\), choose \(x = 1\): \(1(1 - 3) = -2 < 0\)<br />- For \(x > 3\), choose \(x = 4\): \(4(4 - 3) = 4 > 0\)<br /><br />Thus, the solution for this case is \(x \in (-\infty, 0) \cup (3, \infty)\).<br /><br />**Case 2:** \(-(x^2 - x) > 2x\) or equivalently \(x^2 - x < -2x\)<br /><br />This simplifies to:<br />\[ x^2 + x < 0 \]<br />Factor the quadratic:<br />\[ x(x + 1) < 0 \]<br /><br />The critical points are \(x = 0\) and \(x = -1\). We test intervals around these points:<br /><br />- For \(x < -1\), choose \(x = -2\): \((-2)(-2 + 1) = 2 > 0\)<br />- For \(-1 < x < 0\), choose \(x = -0.5\): \((-0.5)(-0.5 + 1) = 0.25 > 0\)<br />- For \(x > 0\), choose \(x = 1\): \(1(1 + 1) = 2 > 0\)<br /><br />Thus, the solution for this case is \(x \in (-1, 0)\).<br /><br />Combining both cases, the solution to the inequality \(|x^2 - x| > 2x\) is:<br />\[ x \in (-\infty, -1) \cup (0, 3) \cup (3, \infty) \]<br /><br />Now, we find the integers that do not satisfy this inequality. These integers are in the intervals where the inequality does not hold, which are:<br />\[ x \in [-1, 0] \cup [3, 3] \]<br /><br />The integers in these intervals are: \(-1, 0, 3\).<br /><br />Therefore, there are 3 integers where the inequality does not apply. The answer is:<br /><br />B) 3