If both roots of the quadratic equation `x^(2)+x+p=0` exceed p where `p epsilon R`, then p must lie in the interval

Options

A. `(-oo,1)`
B. `(-oo,2)`
C. `(-oo,-2)uu(0,1//4)`
D. `(-2,1)`

Answer

To solve the problem, we need to analyze the quadratic equation \( x^2 + x + p = 0 \) and find the conditions under which both roots exceed \( p \).

### Step-by-Step Solution:

1. **Identify the Quadratic Equation**:
The given quadratic equation is:
\[ x^2 + x + p = 0 \]

2. **Condition for Real Roots**:
For the quadratic equation to have real roots, the discriminant must be greater than zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 1 \), and \( c = p \). Thus,
\[ D = 1^2 - 4 \cdot 1 \cdot p = 1 - 4p \]
We need \( D > 0 \):
\[ 1 - 4p > 0 \]
Rearranging gives:
\[ 4p < 1 \quad \Rightarrow \quad p < \frac{1}{4} \]

3. **Condition for Roots Exceeding \( p \)**:
Let the roots of the equation be \( r_1 \) and \( r_2 \). For both roots to exceed \( p \), we can use Vieta's formulas:
- The sum of the roots \( r_1 + r_2 = -b = -1 \)
- The product of the roots \( r_1 r_2 = c = p \)

Since both roots exceed \( p \), we can express this condition as:
\[ f(p) = p^2 + p + p > 0 \]
Simplifying gives:
\[ p^2 + 2p > 0 \]
Factoring out \( p \):
\[ p(p + 2) > 0 \]

4. **Finding the Intervals**:
The inequality \( p(p + 2) > 0 \) holds when:
- \( p < -2 \) or \( p > 0 \)

5. **Combining Conditions**:
We have two conditions:
- From the discriminant: \( p < \frac{1}{4} \)
- From the roots exceeding \( p \): \( p < -2 \) or \( p > 0 \)

We need to find the intersection of these intervals:
- For \( p < -2 \), it is valid since \( -2 < \frac{1}{4} \).
- For \( p > 0 \), it does not satisfy \( p < \frac{1}{4} \).

Thus, the only valid interval for \( p \) is:
\[ p \in (-\infty, -2) \cup (0, \frac{1}{4}) \]

### Final Answer:
The values of \( p \) must lie in the interval:
\[ p \in (-\infty, -2) \cup (0, \frac{1}{4}) \]