IF `sqrt(9x^2 + 6x +1) lt (2-x)` then

Options

A. `((-3)/(2), (1)/(4))`
B. `[(-3)/(2), (1)/(4))`
C. `((-3)/(2), (1)/(4)]`
D. (0, 2)

Answer

To solve the inequality \( \sqrt{9x^2 + 6x + 1} < 2 - x \), we will follow these steps:

### Step 1: Square both sides
To eliminate the square root, we square both sides of the inequality:
\[ 9x^2 + 6x + 1 < (2 - x)^2 \]

### Step 2: Expand the right side
Now, we expand the right side:
\[ (2 - x)^2 = 4 - 4x + x^2 \]
Thus, the inequality becomes:
\[ 9x^2 + 6x + 1 < 4 - 4x + x^2 \]

### Step 3: Rearrange the inequality
Next, we move all terms to one side:
\[ 9x^2 + 6x + 1 - 4 + 4x - x^2 < 0 \]
This simplifies to:
\[ (9x^2 - x^2) + (6x + 4x) + (1 - 4) < 0 \]
\[ 8x^2 + 10x - 3 < 0 \]

### Step 4: Factor the quadratic expression
Now we need to factor the quadratic expression \( 8x^2 + 10x - 3 \). We can do this by finding two numbers that multiply to \( 8 \times (-3) = -24 \) and add to \( 10 \). The numbers are \( 12 \) and \( -2 \). Thus, we can rewrite the expression:
\[ 8x^2 + 12x - 2x - 3 < 0 \]
Now, we group the terms:
\[ 4x(2x + 3) - 1(2x + 3) < 0 \]
Factoring out \( (2x + 3) \):
\[ (2x + 3)(4x - 1) < 0 \]

### Step 5: Find the critical points
To find the critical points, we set each factor to zero:
1. \( 2x + 3 = 0 \) gives \( x = -\frac{3}{2} \)
2. \( 4x - 1 = 0 \) gives \( x = \frac{1}{4} \)

### Step 6: Test intervals
We will now test the intervals determined by the critical points \( x = -\frac{3}{2} \) and \( x = \frac{1}{4} \):
- Interval 1: \( (-\infty, -\frac{3}{2}) \)
- Interval 2: \( (-\frac{3}{2}, \frac{1}{4}) \)
- Interval 3: \( (\frac{1}{4}, \infty) \)

Choose test points from each interval:
1. For \( x = -2 \) (Interval 1): \( (2(-2) + 3)(4(-2) - 1) = (-4 + 3)(-8 - 1) = (-1)(-9) > 0 \)
2. For \( x = 0 \) (Interval 2): \( (2(0) + 3)(4(0) - 1) = (3)(-1) < 0 \)
3. For \( x = 1 \) (Interval 3): \( (2(1) + 3)(4(1) - 1) = (2 + 3)(4 - 1) = (5)(3) > 0 \)

### Step 7: Conclusion
The inequality \( (2x + 3)(4x - 1) < 0 \) holds true in the interval:
\[ x \in \left(-\frac{3}{2}, \frac{1}{4}\right) \]

### Final Answer
Thus, the solution to the inequality \( \sqrt{9x^2 + 6x + 1} < 2 - x \) is:
\[ x \in \left(-\frac{3}{2}, \frac{1}{4}\right) \]