The solution set of the quadratic inequality `3- 2x^(2) le 5x` is

Options

A. (a) `( - oo, -3]`
B. (b) `[1/2 , oo)`
C. (c) `[-3,1/2]`
D. (d) `(- oo,-3] cup [1/2 ,oo)`

Answer

To solve the quadratic inequality \(3 - 2x^2 \leq 5x\), we will follow these steps:

### Step 1: Rearranging the Inequality
Start by rearranging the inequality to one side:
\[ 3 - 2x^2 - 5x \leq 0 \]
This simplifies to:
\[ -2x^2 - 5x + 3 \leq 0 \]

### Step 2: Multiplying by -1
To make the leading coefficient positive, multiply the entire inequality by -1 (remember to reverse the inequality sign):
\[ 2x^2 + 5x - 3 \geq 0 \]

### Step 3: Factoring the Quadratic
Next, we need to factor the quadratic expression \(2x^2 + 5x - 3\). We look for two numbers that multiply to \(2 \times -3 = -6\) and add to \(5\). The numbers \(6\) and \(-1\) work:
\[ 2x^2 + 6x - x - 3 \geq 0 \]
Now, we can group the terms:
\[ (2x^2 + 6x) + (-x - 3) \geq 0 \]
Factoring by grouping:
\[ 2x(x + 3) - 1(x + 3) \geq 0 \]
This gives us:
\[ (2x - 1)(x + 3) \geq 0 \]

### Step 4: Finding the Critical Points
Set each factor to zero to find the critical points:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x + 3 = 0 \Rightarrow x = -3\)

### Step 5: Testing Intervals
Now we test the intervals determined by the critical points \(-3\) and \(\frac{1}{2}\):
- Interval 1: \( (-\infty, -3) \)
- Interval 2: \( (-3, \frac{1}{2}) \)
- Interval 3: \( (\frac{1}{2}, \infty) \)

Choose test points from each interval:
1. For \(x = -4\) in \((- \infty, -3)\):
\[ (2(-4) - 1)(-4 + 3) = (-8 - 1)(-1) = 9 > 0 \]
2. For \(x = 0\) in \((-3, \frac{1}{2})\):
\[ (2(0) - 1)(0 + 3) = (-1)(3) = -3 < 0 \]
3. For \(x = 1\) in \((\frac{1}{2}, \infty)\):
\[ (2(1) - 1)(1 + 3) = (2 - 1)(4) = 4 > 0 \]

### Step 6: Writing the Solution Set
From the tests, we find that the inequality is satisfied in the intervals:
- \( (-\infty, -3] \) (includes -3)
- \( [\frac{1}{2}, \infty) \) (includes \(\frac{1}{2}\))

Thus, the solution set is:
\[ (-\infty, -3] \cup [\frac{1}{2}, \infty) \]

### Final Answer
The solution set of the quadratic inequality \(3 - 2x^2 \leq 5x\) is:
\[ \boxed{(-\infty, -3] \cup [\frac{1}{2}, \infty)} \]