If `3 - 2x^(2) gt 5x, x in R ` , then the set of solutions is

Options

A. A. `(- oo,-1/2)`
B. B. `(3,oo)`
C. C. `(-1/2,3)`
D. D. `[-1/2,3]`

Answer

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

### Step 1: Rearranging the Inequality
Start by rearranging the inequality to one side:
\[ 3 - 2x^2 - 5x > 0 \]
This can be rewritten as:
\[ -2x^2 - 5x + 3 > 0 \]

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

### Step 3: Factoring the Quadratic
Next, we need to factor the quadratic expression \(2x^2 + 5x - 3\). We can do this by finding two numbers that multiply to \(2 \times -3 = -6\) and add to \(5\). These numbers are \(6\) and \(-1\).

Now, we can rewrite \(5x\) as \(6x - x\):
\[ 2x^2 + 6x - x - 3 < 0 \]
Grouping the terms:
\[ (2x^2 + 6x) + (-x - 3) < 0 \]
Factoring by grouping:
\[ 2x(x + 3) - 1(x + 3) < 0 \]
Factoring out \((x + 3)\):
\[ (2x - 1)(x + 3) < 0 \]

### 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 will test the intervals determined by the critical points \(x = -3\) and \(x = \frac{1}{2}\):
1. **Interval 1:** \( (-\infty, -3) \)
- Choose \(x = -4\): \((2(-4) - 1)(-4 + 3) = (-8 - 1)(-1) = 9 > 0\)
2. **Interval 2:** \( (-3, \frac{1}{2}) \)
- Choose \(x = 0\): \((2(0) - 1)(0 + 3) = (-1)(3) = -3 < 0\)
3. **Interval 3:** \( (\frac{1}{2}, \infty) \)
- Choose \(x = 1\): \((2(1) - 1)(1 + 3) = (2 - 1)(4) = 4 > 0\)

### Step 6: Writing the Solution
The inequality \( (2x - 1)(x + 3) < 0 \) holds true in the interval:
\[ (-3, \frac{1}{2}) \]

### Final Solution
Thus, the set of solutions is:
\[ x \in (-3, \frac{1}{2}) \]