If -`x^(2) + 3x + 4 gt 0` . then which one of the following is correct?

Options

A. `x in (-1,4)`
B. `x in [-1,4]`
C. `x in (infty ,-1) mu(4,infty)`
D. `x in (-infty,-1) mu (4,infty)`

Answer

To solve the inequality \(-x^2 + 3x + 4 > 0\), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ -x^2 + 3x + 4 > 0 \]
To make it easier to work with, we can multiply the entire inequality by \(-1\). Remember that when we multiply or divide an inequality by a negative number, we must reverse the inequality sign:
\[ x^2 - 3x - 4 < 0 \]

### Step 2: Factor the Quadratic
Next, we need to factor the quadratic expression \(x^2 - 3x - 4\). We look for two numbers that multiply to \(-4\) (the constant term) and add to \(-3\) (the coefficient of \(x\)). The numbers \(-4\) and \(1\) work:
\[ x^2 - 4x + 1x - 4 = (x - 4)(x + 1) < 0 \]

### Step 3: Find the Roots
Now we set the factors equal to zero to find the roots:
\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
So, the roots are \(x = 4\) and \(x = -1\).

### Step 4: Test Intervals
Next, we will test the intervals determined by the roots \(-1\) and \(4\):
1. **Interval 1:** \( (-\infty, -1) \)
2. **Interval 2:** \( (-1, 4) \)
3. **Interval 3:** \( (4, \infty) \)

We will choose test points from each interval to determine where the inequality holds true.

- **For Interval 1**: Choose \(x = -2\):
\[ (-2 - 4)(-2 + 1) = (-6)(-1) = 6 > 0 \quad \text{(not valid)} \]

- **For Interval 2**: Choose \(x = 0\):
\[ (0 - 4)(0 + 1) = (-4)(1) = -4 < 0 \quad \text{(valid)} \]

- **For Interval 3**: Choose \(x = 5\):
\[ (5 - 4)(5 + 1) = (1)(6) = 6 > 0 \quad \text{(not valid)} \]

### Step 5: Conclusion
The inequality \( (x - 4)(x + 1) < 0 \) holds true in the interval:
\[ (-1, 4) \]
Thus, the solution to the inequality \(-x^2 + 3x + 4 > 0\) is:
\[ x \in (-1, 4) \]