The solution set of `x^(2) + 2 le 3x le 2x^(2)-5`, is

Options

A. `phi`
B. `[1, 2]`
C. `(-oo, -1] cup [5//2, oo)`
D. none of these

Answer

To solve the inequality \( x^2 + 2 \leq 3x \leq 2x^2 - 5 \), we will break it down into two parts and solve each inequality separately.

### Step 1: Solve \( x^2 + 2 \leq 3x \)

1. Rearrange the inequality:
\[ x^2 - 3x + 2 \leq 0 \]

2. Factor the quadratic:
\[ (x - 1)(x - 2) \leq 0 \]

3. Determine the critical points:
The critical points are \( x = 1 \) and \( x = 2 \).

4. Test intervals:
- For \( x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 - 2) = 2 > 0 \) (not in solution set)
- For \( 1 \leq x \leq 2 \) (e.g., \( x = 1.5 \)): \( (1.5 - 1)(1.5 - 2) = -0.25 < 0 \) (in solution set)
- For \( x > 2 \) (e.g., \( x = 3 \)): \( (3 - 1)(3 - 2) = 2 > 0 \) (not in solution set)

5. Conclusion for this part:
\[ 1 \leq x \leq 2 \]

### Step 2: Solve \( 3x \leq 2x^2 - 5 \)

1. Rearrange the inequality:
\[ 2x^2 - 3x - 5 \geq 0 \]

2. Factor the quadratic:
To factor, we can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \]
\[ = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm 7}{4} \]
Thus, the roots are:
\[ x = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1 \]

3. Rewrite the inequality using the roots:
\[ (x - \frac{5}{2})(x + 1) \geq 0 \]

4. Determine the intervals:
- For \( x < -1 \) (e.g., \( x = -2 \)): \( (-2 - \frac{5}{2})(-2 + 1) = \text{positive} \) (in solution set)
- For \( -1 < x < \frac{5}{2} \) (e.g., \( x = 0 \)): \( (0 - \frac{5}{2})(0 + 1) = \text{negative} \) (not in solution set)
- For \( x > \frac{5}{2} \) (e.g., \( x = 3 \)): \( (3 - \frac{5}{2})(3 + 1) = \text{positive} \) (in solution set)

5. Conclusion for this part:
\[ x \leq -1 \quad \text{or} \quad x \geq \frac{5}{2} \]

### Step 3: Solve \( x^2 + 2 \leq 2x^2 - 5 \)

1. Rearrange the inequality:
\[ x^2 - 7 \leq 0 \]

2. Factor:
\[ (x - \sqrt{7})(x + \sqrt{7}) \leq 0 \]

3. Determine the critical points:
The critical points are \( x = -\sqrt{7} \) and \( x = \sqrt{7} \).

4. Test intervals:
- For \( x < -\sqrt{7} \): (e.g., \( x = -3 \)): positive (not in solution set)
- For \( -\sqrt{7} \leq x \leq \sqrt{7} \): (e.g., \( x = 0 \)): negative (in solution set)
- For \( x > \sqrt{7} \): positive (not in solution set)

5. Conclusion for this part:
\[ -\sqrt{7} \leq x \leq \sqrt{7} \]

### Final Step: Combine the results

1. From the first inequality, we have \( 1 \leq x \leq 2 \).
2. From the second inequality, we have \( x \leq -1 \) or \( x \geq \frac{5}{2} \).
3. From the third inequality, we have \( -\sqrt{7} \leq x \leq \sqrt{7} \).

### Conclusion:

The solution set is the intersection of all three parts. Since there are no common values in the ranges \( [1, 2] \), \( (-\infty, -1] \cup [\frac{5}{2}, \infty) \), and \( [-\sqrt{7}, \sqrt{7}] \), we conclude that there is no solution.