When solving \( \boldsymbol{x} \) in the equation \( \boldsymbol{x}:-2 x^{2}+2 \geq-3 x \), the answer is \( \left\{x \left\lvert\,-\frac{1}{2} \leq x \geq 2\right.\right\} \) Select one: True False Solve the following inequality: \( x^{2} \geq-x+6 \) a. \( \{x \mid x \geq-3 \) or \( x \leq 2\} \) b. \( \{x \mid-3>x<2\} \) c. \( \{x \mid-3 \leq x \leq 2\} \) d. \( \{x \mid x \leq-3 \) or \( x \geq 2\} \) \( 12-2 x^{2}-y \leq 0 \) is a quadratic inequality. Select one:

Let's break down the problem step by step. ### Part 1: Analyzing the first inequality The inequality given is: \[ -2x^2 + 2 \geq -3x \] First, we can rearrange this inequality to standard form: \[ -2x^2 + 3x + 2 \geq 0 \] Next, we can multiply through by -1 (remembering to reverse the inequality): \[ 2x^2 - 3x - 2 \leq 0 \] Now, we will solve the quadratic equation \(2x^2 - 3x - 2 = 0\) to find the critical points. We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 2\), \(b = -3\), and \(c = -2\). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4(2)(-2) = 9 + 16 = 25 \] Now, substituting into the quadratic formula: \[ x = \frac{3 \pm \sqrt{25}}{2 \cdot 2} = \frac{3 \pm 5}{4} \] This gives us two solutions: \[ x_1 = \frac{8}{4} = 2, \quad x_2 = \frac{-2}{4} = -\frac{1}{2} \] Now we have the critical points \(x = -\frac{1}{2}\) and \(x = 2\). Next, we will test intervals around these critical points to determine where the inequality holds: 1. **Interval \( (-\infty, -\frac{1}{2}) \)**: Choose \(x = -1\): \[ 2(-1)^2 - 3(-1) - 2 = 2 + 3 - 2 = 3 \quad (\text{True}) \] 2. **Interval \( (-\frac{1}{2}, 2) \)**: Choose \(x = 0\): \[ 2(0)^2 - 3(0) - 2 = -2 \quad (\text{False}) \] 3. **Interval \( (2, \infty) \)**: Choose \(x = 3\): \[ 2(3)^2 - 3(3) - 2 = 18 - 9 - 2 = 7 \quad (\text{True}) \] Thus, the solution to the inequality \(2x^2 - 3x - 2 \leq 0\) is: \[ x \in \left[-\frac{1}{2}, 2\right] \] This means the statement \( \left\{x \mid -\frac{1}{2} \leq x \geq 2\right\} \) is **False**. ### Part 2: Solving the second inequality Now, we need to solve the inequality: \[ x^2 \geq -x + 6 \] Rearranging gives: \[ x^2 + x - 6 \geq 0 \] Next, we factor the quadratic: \[ (x + 3)(x - 2) \geq 0 \] The critical points are \(x = -3\) and \(x = 2\). We will test intervals around these points: 1. **Interval \( (-\infty, -3) \)**: Choose \(x = -4\): \[ (-4 + 3)(-4 - 2) = (-1)(-6) = 6 \quad (\text{True}) \] 2. **Interval \( (-3, 2) \)**: Choose \(x = 0\): \[ (0 + 3)(0 - 2) = (3)(-2) = -6 \quad (\text{False}) \] 3. **Interval \( (2, \infty) \)**: Choose \(x = 3\): \[ (3 + 3)(3 - 2) = (6)(1) = 6 \quad (\text{True}) \] Thus, the solution to the inequality \(x^2 + x - 6 \geq 0\) is: \[ x \in (-\infty, -3] \cup [2, \infty) \] This corresponds to option **d**: \( \{x \mid x \leq -3 \text{ or } x \geq 2\} \). ### Part 3: Analyzing the quadratic inequality The inequality \(12 - 2x^2 - y \leq 0\) can be rearranged to: \[ y \geq 12 - 2x^2 \] This represents a downward-opening parabola in the xy-plane. The statement that it is a quadratic inequality is **True**. ### Summary of Answers: 1. The first statement is **False**. 2. The correct option for the second inequality is **d**. 3. The statement about the quadratic inequality is **True**.