If x is real and find the limits between which x must lie , when `|(x^(2) - 3x-1)/(x^(2) + x+1)| lt 3`

Options

A. All values of x
B. `-2 lt x lt -1`
C. `(-oo, 2) cup (4, oo)`
D. `(-oo, -2) cup (-1, oo)`

Answer

To solve the inequality \( \left| \frac{x^2 - 3x - 1}{x^2 + x + 1} \right| < 3 \), we will break it down into two separate inequalities.

### Step 1: Setting up the inequalities
We can rewrite the absolute value inequality as two separate inequalities:
1. \( \frac{x^2 - 3x - 1}{x^2 + x + 1} < 3 \)
2. \( \frac{x^2 - 3x - 1}{x^2 + x + 1} > -3 \)

### Step 2: Solve the first inequality
For the first inequality:
\[ \frac{x^2 - 3x - 1}{x^2 + x + 1} < 3 \]
Multiply both sides by \( x^2 + x + 1 \) (which is always positive):
\[ x^2 - 3x - 1 < 3(x^2 + x + 1) \]
Expanding the right side:
\[ x^2 - 3x - 1 < 3x^2 + 3x + 3 \]
Rearranging gives:
\[ 0 < 3x^2 + 3x + 3 - x^2 + 3x + 1 \]
This simplifies to:
\[ 0 < 2x^2 + 6x + 4 \]
Dividing through by 2:
\[ 0 < x^2 + 3x + 2 \]
Factoring the quadratic:
\[ 0 < (x + 1)(x + 2) \]
The critical points are \( x = -1 \) and \( x = -2 \).

### Step 3: Test intervals for the first inequality
We test the intervals:
- For \( x < -2 \): Choose \( x = -3 \) → \( (-3 + 1)(-3 + 2) = (-2)(-1) > 0 \) (True)
- For \( -2 < x < -1 \): Choose \( x = -1.5 \) → \( (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) < 0 \) (False)
- For \( x > -1 \): Choose \( x = 0 \) → \( (0 + 1)(0 + 2) = (1)(2) > 0 \) (True)

Thus, the solution for the first inequality is:
\[ x \in (-\infty, -2) \cup (-1, \infty) \]

### Step 4: Solve the second inequality
Now we solve the second inequality:
\[ \frac{x^2 - 3x - 1}{x^2 + x + 1} > -3 \]
Multiply both sides by \( x^2 + x + 1 \):
\[ x^2 - 3x - 1 > -3(x^2 + x + 1) \]
Expanding the right side:
\[ x^2 - 3x - 1 > -3x^2 - 3x - 3 \]
Rearranging gives:
\[ 0 > -3x^2 - 3x - 3 - x^2 + 3x + 1 \]
This simplifies to:
\[ 0 > -4x^2 - 2 \]
Dividing through by -1 (reversing the inequality):
\[ 0 < 4x^2 + 2 \]
This is always true since \( 4x^2 + 2 > 0 \) for all real \( x \).

### Step 5: Combine the results
The solution to the second inequality is all real numbers.

### Final Solution
The combined solution from both inequalities is:
\[ x \in (-\infty, -2) \cup (-1, \infty) \]