\( 1.3 \quad 2 x ^ { 2 } - 2 \geq 3 x \)

To solve the inequality \(2x^2 - 2 \geq 3x\), first, we can rearrange it to form a standard quadratic inequality. Move all terms to one side: \[2x^2 - 3x - 2 \geq 0\] Now, factor the quadratic expression. You want factors of \(-4\) (the product of \(2\) and \(-2\)) that add up to \(-3\). The factors \(-4\) and \(1\) work, so we can rewrite it as: \[(2x + 1)(x - 2) \geq 0\] Next, determine the roots by setting each factor to zero: 1. \(2x + 1 = 0 \Rightarrow x = -\frac{1}{2}\) 2. \(x - 2 = 0 \Rightarrow x = 2\) Now, we have critical points at \(x = -\frac{1}{2}\) and \(x = 2\). To solve the inequality, test intervals defined by these critical points: 1. \( (-\infty, -\frac{1}{2}) \) 2. \( (-\frac{1}{2}, 2) \) 3. \( (2, \infty) \) Choose test points from each interval and evaluate the inequality: - For \(x = -1\) (first interval): \( (2(-1) + 1)(-1 - 2) = (-1)(-3) = 3 \geq 0\) (True) - For \(x = 0\) (second interval): \( (2(0) + 1)(0 - 2) = (1)(-2) = -2 \geq 0\) (False) - For \(x = 3\) (third interval): \( (2(3) + 1)(3 - 2) = (7)(1) = 7 \geq 0\) (True) The solution set includes the intervals where the inequality holds and the roots themselves since the inequality is non-strict (≥0). Therefore, the solution is: \[ x \in \left(-\infty, -\frac{1}{2}\right] \cup [2, \infty) \]