If the absolute value of the difference of the roots of the equation `x^(2) + ax + 1 = 0 "exceeds" sqrt(3a)`, then

Options

A. `a in (-oo, -1) uu (4, oo)`
B. `a in (4, oo)`
C. `a in (-1, 4)`
D. `a in [ 0, 4)`

Answer

To solve the problem, we need to find the range of \( a \) such that the absolute value of the difference of the roots of the equation \( x^2 + ax + 1 = 0 \) exceeds \( \sqrt{3a} \).

### Step-by-Step Solution:

1. **Identify the Roots**:
For the quadratic equation \( x^2 + ax + 1 = 0 \), we can denote the roots as \( \alpha \) and \( \beta \).

2. **Use the Formula for the Difference of Roots**:
The difference of the roots can be expressed as:
\[ |\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} \]
From Vieta's formulas, we know:
- \( \alpha + \beta = -\frac{b}{a} = -a \)
- \( \alpha \beta = \frac{c}{a} = 1 \)

Substituting these into the formula gives:
\[ |\alpha - \beta| = \sqrt{(-a)^2 - 4 \cdot 1} = \sqrt{a^2 - 4} \]

3. **Set Up the Inequality**:
We need to find when this difference exceeds \( \sqrt{3a} \):
\[ \sqrt{a^2 - 4} > \sqrt{3a} \]

4. **Square Both Sides**:
To eliminate the square roots, we square both sides:
\[ a^2 - 4 > 3a \]

5. **Rearrange the Inequality**:
Rearranging gives:
\[ a^2 - 3a - 4 > 0 \]

6. **Factor the Quadratic**:
We can factor the quadratic:
\[ (a - 4)(a + 1) > 0 \]

7. **Determine the Intervals**:
To find the intervals where this inequality holds, we can test the intervals determined by the roots \( a = -1 \) and \( a = 4 \):
- For \( a < -1 \): Choose \( a = -2 \) → \( (-2 - 4)(-2 + 1) = (-6)(-1) > 0 \) (True)
- For \( -1 < a < 4 \): Choose \( a = 0 \) → \( (0 - 4)(0 + 1) = (-4)(1) < 0 \) (False)
- For \( a > 4 \): Choose \( a = 5 \) → \( (5 - 4)(5 + 1) = (1)(6) > 0 \) (True)

8. **Conclusion**:
The solution to the inequality \( (a - 4)(a + 1) > 0 \) is:
\[ a \in (-\infty, -1) \cup (4, \infty) \]

### Final Answer:
The values of \( a \) for which the absolute value of the difference of the roots exceeds \( \sqrt{3a} \) are:
\[ a \in (-\infty, -1) \cup (4, \infty) \]