The range of values of x which satisfy `2x^(2) + 9x + 4 lt 0` and `x^(2) - 5x + 6 lt 0` is

Options

A. `(-2,-1)`
B. `((1)/(2), 4)`
C. `(2, 3)`
D. `phi`

Answer

To solve the inequalities \(2x^2 + 9x + 4 < 0\) and \(x^2 - 5x + 6 < 0\), we will follow these steps:

### Step 1: Solve the first inequality \(2x^2 + 9x + 4 < 0\)

1. **Factor the quadratic**:
We need to factor \(2x^2 + 9x + 4\). We look for two numbers that multiply to \(2 \times 4 = 8\) and add to \(9\). The numbers \(8\) and \(1\) work.
\[ 2x^2 + 8x + 1x + 4 < 0 \]
Rearranging gives:
\[ (2x^2 + 8x) + (1x + 4) < 0 \]
Factoring by grouping:
\[ 2x(x + 4) + 1(x + 4) < 0 \]
This simplifies to:
\[ (2x + 1)(x + 4) < 0 \]

2. **Find the roots**:
Set each factor to zero:
\[ 2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2} \]
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]

3. **Test intervals**:
The critical points are \(x = -4\) and \(x = -\frac{1}{2}\). We test intervals:
- For \(x < -4\) (e.g., \(x = -5\)): \((2(-5) + 1)(-5 + 4) = (-10 + 1)(-1) = 9 > 0\)
- For \(-4 < x < -\frac{1}{2}\) (e.g., \(x = -1\)): \((2(-1) + 1)(-1 + 4) = (-2 + 1)(3) = -3 < 0\)
- For \(x > -\frac{1}{2}\) (e.g., \(x = 0\)): \((2(0) + 1)(0 + 4) = (1)(4) = 4 > 0\)

4. **Solution for the first inequality**:
The solution is:
\[ -4 < x < -\frac{1}{2} \]

### Step 2: Solve the second inequality \(x^2 - 5x + 6 < 0\)

1. **Factor the quadratic**:
We need to factor \(x^2 - 5x + 6\). The numbers \(2\) and \(3\) work.
\[ (x - 2)(x - 3) < 0 \]

2. **Find the roots**:
Set each factor to zero:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

3. **Test intervals**:
The critical points are \(x = 2\) and \(x = 3\). We test intervals:
- For \(x < 2\) (e.g., \(x = 0\)): \((0 - 2)(0 - 3) = ( -2)( -3) = 6 > 0\)
- For \(2 < x < 3\) (e.g., \(x = 2.5\)): \((2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0\)
- For \(x > 3\) (e.g., \(x = 4\)): \((4 - 2)(4 - 3) = (2)(1) = 2 > 0\)

4. **Solution for the second inequality**:
The solution is:
\[ 2 < x < 3 \]

### Step 3: Find the intersection of the solutions

- The first inequality gives us \(-4 < x < -\frac{1}{2}\).
- The second inequality gives us \(2 < x < 3\).

### Conclusion

Since there are no common values between the two intervals \((-4, -\frac{1}{2})\) and \((2, 3)\), the solution set for the combined inequalities is the null set.

### Final Answer:
The range of values of \(x\) that satisfy both inequalities is the null set.

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