Find all the values of x that satisfy the following inequation `|(x^2-3x-1)/(x^2+x+1)|lt3`

Options

A. `(-oo,-1)uu(-2,oo)`
B. `(-oo,-2)uu(-1,oo)`
C. `(-oo,oo)`
D. `(-9,6)`

Answer

To solve the inequation \(\left|\frac{x^2 - 3x - 1}{x^2 + x + 1}\right| < 3\), we will follow these steps:

### Step 1: Rewrite the Inequation
We start by rewriting the inequation without the absolute value:
\[ -\frac{x^2 - 3x - 1}{x^2 + x + 1} < 3 < \frac{x^2 - 3x - 1}{x^2 + x + 1} \]

### Step 2: Solve the Right Side of the Inequation
First, we solve:
\[ \frac{x^2 - 3x - 1}{x^2 + x + 1} < 3 \]
This can be rewritten as:
\[ x^2 - 3x - 1 < 3(x^2 + x + 1) \]
Expanding the right side:
\[ x^2 - 3x - 1 < 3x^2 + 3x + 3 \]
Rearranging gives:
\[ 0 < 2x^2 + 6x + 4 \]
Dividing the entire inequality by 2:
\[ 0 < x^2 + 3x + 2 \]
Factoring:
\[ 0 < (x + 1)(x + 2) \]
The critical points are \(x = -1\) and \(x = -2\). We analyze the sign of the expression:
- For \(x < -2\), both factors are negative, so the product is positive.
- For \(-2 < x < -1\), one factor is negative and the other is positive, so the product is negative.
- For \(x > -1\), both factors are positive, so the product is positive.

Thus, the solution to this part is:
\[ x < -2 \quad \text{or} \quad x > -1 \]

### Step 3: Solve the Left Side of the Inequation
Now we solve:
\[ -\frac{x^2 - 3x - 1}{x^2 + x + 1} < 3 \]
This can be rewritten as:
\[ \frac{x^2 - 3x - 1}{x^2 + x + 1} > -3 \]
Rearranging gives:
\[ x^2 - 3x - 1 > -3(x^2 + x + 1) \]
Expanding the right side:
\[ x^2 - 3x - 1 > -3x^2 - 3x - 3 \]
Rearranging gives:
\[ 4x^2 > -2 \]
This is always true since \(4x^2\) is always non-negative. Therefore, this part does not impose any additional restrictions.

### Step 4: Combine the Results
From Step 2, we have:
\[ x < -2 \quad \text{or} \quad x > -1 \]
From Step 3, we have no additional restrictions.

### Step 5: Exclude Points Where the Denominator is Zero
The denominator \(x^2 + x + 1\) is never zero for real \(x\) since its discriminant \(b^2 - 4ac = 1 - 4 = -3\) is negative.

### Final Solution
Thus, the final solution is:
\[ x < -2 \quad \text{or} \quad x > -1 \]