The set of real values of `x` for which `(10x^2 +17x-34)/(x^2 + 2x - 3) < 8,` is

Options

A. `(-5//2, 2)`
B. `(0, oo)`
C. `(-1, oo)`
D. none of these

Answer

To solve the inequality \(\frac{10x^2 + 17x - 34}{x^2 + 2x - 3} < 8\), we will follow these steps:

### Step 1: Rearranging the Inequality
First, we will move 8 to the left-hand side of the inequality:
\[ \frac{10x^2 + 17x - 34}{x^2 + 2x - 3} - 8 < 0 \]

### Step 2: Finding a Common Denominator
The common denominator is \(x^2 + 2x - 3\). We rewrite the inequality:
\[ \frac{10x^2 + 17x - 34 - 8(x^2 + 2x - 3)}{x^2 + 2x - 3} < 0 \]

### Step 3: Simplifying the Numerator
Now we simplify the numerator:
\[ 10x^2 + 17x - 34 - 8x^2 - 16x + 24 < 0 \]
Combining like terms gives:
\[ (10x^2 - 8x^2) + (17x - 16x) + (-34 + 24) < 0 \]
\[ 2x^2 + x - 10 < 0 \]

### Step 4: Factoring the Quadratic
Next, we will factor the quadratic \(2x^2 + x - 10\). We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-10)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 1 + 80 = 81 \quad \Rightarrow \quad \sqrt{81} = 9 \]
Thus, the roots are:
\[ x = \frac{-1 + 9}{4} = 2 \quad \text{and} \quad x = \frac{-1 - 9}{4} = -2.5 \]

### Step 5: Finding Critical Points
The critical points are \(x = -2.5\) and \(x = 2\). We also need to consider the denominator \(x^2 + 2x - 3\), which factors to \((x - 1)(x + 3)\). The critical points from the denominator are \(x = 1\) and \(x = -3\).

### Step 6: Testing Intervals
We will test the intervals determined by the critical points:
- \((-∞, -3)\)
- \((-3, -2.5)\)
- \((-2.5, 1)\)
- \((1, 2)\)
- \((2, ∞)\)

1. **Interval \((-∞, -3)\)**: Choose \(x = -4\):
\[ \frac{2(-4)^2 + (-4) - 10}{(-4)^2 + 2(-4) - 3} = \frac{32 - 4 - 10}{16 - 8 - 3} = \frac{18}{5} > 0 \]

2. **Interval \((-3, -2.5)\)**: Choose \(x = -2.7\):
\[ \frac{2(-2.7)^2 + (-2.7) - 10}{(-2.7)^2 + 2(-2.7) - 3} < 0 \]

3. **Interval \((-2.5, 1)\)**: Choose \(x = 0\):
\[ \frac{2(0)^2 + (0) - 10}{(0)^2 + 2(0) - 3} < 0 \]

4. **Interval \((1, 2)\)**: Choose \(x = 1.5\):
\[ \frac{2(1.5)^2 + (1.5) - 10}{(1.5)^2 + 2(1.5) - 3} < 0 \]

5. **Interval \((2, ∞)\)**: Choose \(x = 3\):
\[ \frac{2(3)^2 + (3) - 10}{(3)^2 + 2(3) - 3} > 0 \]

### Step 7: Conclusion
The intervals where the inequality is satisfied (less than 0) are:
- \((-3, -2.5)\)
- \((-2.5, 1)\)
- \((1, 2)\)

Thus, the solution set is:
\[ x \in (-3, -2.5) \cup (-2.5, 1) \cup (1, 2) \]