Solve `(6x^2-5x-3)/(x^2-2x+6)le 4`

Explanation

To solve the inequality \(\frac{6x^2 - 5x - 3}{x^2 - 2x + 6} \leq 4\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \frac{6x^2 - 5x - 3}{x^2 - 2x + 6} \leq 4 \]
Subtract 4 from both sides:
\[ \frac{6x^2 - 5x - 3}{x^2 - 2x + 6} - 4 \leq 0 \]

### Step 2: Combine the Terms
To combine the terms, we need a common denominator:
\[ \frac{6x^2 - 5x - 3 - 4(x^2 - 2x + 6)}{x^2 - 2x + 6} \leq 0 \]
Distributing the -4:
\[ \frac{6x^2 - 5x - 3 - 4x^2 + 8x - 24}{x^2 - 2x + 6} \leq 0 \]

### Step 3: Simplify the Numerator
Combine like terms in the numerator:
\[ \frac{(6x^2 - 4x^2) + (-5x + 8x) + (-3 - 24)}{x^2 - 2x + 6} \leq 0 \]
This simplifies to:
\[ \frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \leq 0 \]

### Step 4: Factor the Numerator
Next, we will factor the numerator \(2x^2 + 3x - 27\). We can use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 2\), \(b = 3\), and \(c = -27\).
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-27)}}{2 \cdot 2} \]
\[ x = \frac{-3 \pm \sqrt{9 + 216}}{4} \]
\[ x = \frac{-3 \pm \sqrt{225}}{4} \]
\[ x = \frac{-3 \pm 15}{4} \]
Calculating the two roots:
\[ x_1 = \frac{12}{4} = 3, \quad x_2 = \frac{-18}{4} = -\frac{9}{2} \]

### Step 5: Analyze the Denominator
Now, we analyze the denominator \(x^2 - 2x + 6\). The discriminant is:
\[ (-2)^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20 \]
Since the discriminant is negative, the denominator is always positive.

### Step 6: Set Up the Sign Chart
Now we can set up a sign chart for the numerator \(2x^2 + 3x - 27\):
- The roots are \(x = -\frac{9}{2}\) and \(x = 3\).
- Test intervals: \((- \infty, -\frac{9}{2})\), \((- \frac{9}{2}, 3)\), and \((3, \infty)\).

1. For \(x < -\frac{9}{2}\), choose \(x = -5\):
\[ 2(-5)^2 + 3(-5) - 27 = 50 - 15 - 27 = 8 \quad (\text{positive}) \]
2. For \(-\frac{9}{2} < x < 3\), choose \(x = 0\):
\[ 2(0)^2 + 3(0) - 27 = -27 \quad (\text{negative}) \]
3. For \(x > 3\), choose \(x = 4\):
\[ 2(4)^2 + 3(4) - 27 = 32 + 12 - 27 = 17 \quad (\text{positive}) \]

### Step 7: Determine the Solution Set
The function \(\frac{2x^2 + 3x - 27}{x^2 - 2x + 6} \leq 0\) is satisfied in the interval:
\[ -\frac{9}{2} \leq x \leq 3 \]

### Final Answer
Thus, the solution to the inequality is:
\[ x \in \left[-\frac{9}{2}, 3\right] \]