The solution set of the inequality `log_(5/8)(2x^(2)-x-3/8) ge1` is-

Options

A. `[-1/2, -1/4] cup (3/4, 1)`
B. `[-1/2, 1]`
C. `[-1/2, 1/4] cup (3/4, 1)`
D. `(-infty, -1/4] cup [3/4, infty)`

Answer

To solve the inequality \( \log_{(5/8)}\left(\frac{2x^2 - x - 3}{8}\right) \geq 1 \), we will follow these steps:

### Step 1: Understand the logarithmic inequality
By the definition of logarithms, we know that:
\[ \log_{(5/8)}(a) \geq 1 \implies a \leq (5/8)^1 = \frac{5}{8} \]
Thus, we need to solve:
\[ \frac{2x^2 - x - 3}{8} \leq \frac{5}{8} \]

### Step 2: Eliminate the denominator
Multiplying both sides by 8 (since 8 is positive, the inequality sign remains unchanged):
\[ 2x^2 - x - 3 \leq 5 \]

### Step 3: Rearrange the inequality
Rearranging gives:
\[ 2x^2 - x - 8 \leq 0 \]

### Step 4: Factor the quadratic
To factor \( 2x^2 - x - 8 \), we look for two numbers that multiply to \( 2 \times -8 = -16 \) and add to \(-1\). The numbers are \( -4 \) and \( 2 \):
\[ 2x^2 - 4x + 2x - 8 \leq 0 \]
Factoring by grouping:
\[ 2x(x - 2) + 4(x - 2) \leq 0 \]
This simplifies to:
\[ (2x + 4)(x - 2) \leq 0 \]

### Step 5: Find critical points
Setting each factor to zero gives:
\[ 2x + 4 = 0 \implies x = -2 \]
\[ x - 2 = 0 \implies x = 2 \]
Thus, the critical points are \( x = -2 \) and \( x = 2 \).

### Step 6: Test intervals
We will test the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \).

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (2(-3) + 4)(-3 - 2) = (-6 + 4)(-5) = (-2)(-5) = 10 \quad (\text{positive}) \]

- For \( -2 < x < 2 \) (e.g., \( x = 0 \)):
\[ (2(0) + 4)(0 - 2) = (4)(-2) = -8 \quad (\text{negative}) \]

- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (2(3) + 4)(3 - 2) = (6 + 4)(1) = 10 \quad (\text{positive}) \]

### Step 7: Determine the solution set
The inequality \( (2x + 4)(x - 2) \leq 0 \) is satisfied in the interval:
\[ [-2, 2] \]

### Step 8: Consider the logarithm's domain
We also need to ensure that the argument of the logarithm is positive:
\[ \frac{2x^2 - x - 3}{8} > 0 \implies 2x^2 - x - 3 > 0 \]
Factoring gives:
\[ (2x + 3)(x - 1) > 0 \]
The critical points are \( x = -\frac{3}{2} \) and \( x = 1 \). Testing intervals gives:
- For \( x < -\frac{3}{2} \): positive
- For \( -\frac{3}{2} < x < 1 \): negative
- For \( x > 1 \): positive

Thus, the solution set for \( 2x^2 - x - 3 > 0 \) is:
\[ (-\infty, -\frac{3}{2}) \cup (1, \infty) \]

### Step 9: Combine the intervals
The final solution set is the intersection of:
1. \( [-2, 2] \) from the logarithmic inequality
2. \( (-\infty, -\frac{3}{2}) \cup (1, \infty) \) from the domain of the logarithm

The intersection gives:
\[ [-2, -\frac{3}{2}) \cup (1, 2] \]

### Final Answer
The solution set of the inequality is:
\[ [-2, -\frac{3}{2}) \cup (1, 2] \]