To solve the quadratic equation \(x^2 + 2(a-1)x + (a+5) = 0\), we will analyze the conditions under which the roots are positive, of opposite signs, or negative.
### Step 1: Identify the coefficients
The given quadratic equation can be expressed in the standard form \(ax^2 + bx + c = 0\) where:
- \(a = 1\)
- \(b = 2(a - 1)\)
- \(c = a + 5\)
### Step 2: Determine the condition for real roots
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[
D = b^2 - 4ac \geq 0
\]
Substituting the values of \(a\), \(b\), and \(c\):
\[
D = [2(a - 1)]^2 - 4(1)(a + 5) \geq 0
\]
Calculating \(D\):
\[
D = 4(a - 1)^2 - 4(a + 5)
\]
\[
D = 4[(a - 1)^2 - (a + 5)]
\]
Expanding:
\[
D = 4[a^2 - 2a + 1 - a - 5] = 4[a^2 - 3a - 4]
\]
Setting the discriminant greater than or equal to zero:
\[
a^2 - 3a - 4 \geq 0
\]
### Step 3: Factor the quadratic
To factor \(a^2 - 3a - 4\):
\[
a^2 - 3a - 4 = (a - 4)(a + 1) \geq 0
\]
### Step 4: Determine the intervals
To find the intervals where the product is non-negative, we analyze the critical points \(a = -1\) and \(a = 4\). We test intervals:
- For \(a < -1\): Both factors are negative, so the product is positive.
- For \(-1 < a < 4\): One factor is negative and one is positive, so the product is negative.
- For \(a > 4\): Both factors are positive, so the product is positive.
Thus, the intervals for real roots are:
\[
a \in (-\infty, -1] \cup [4, \infty)
\]
### Step 5: Analyze the roots for positivity, negativity, and opposite signs
1. **Positive Roots**: For both roots to be positive:
- The sum of roots \( \alpha + \beta = -\frac{b}{a} = -2(a - 1) > 0 \Rightarrow a < 1\)
- The product of roots \( \alpha \beta = \frac{c}{a} = a + 5 > 0 \Rightarrow a > -5\)
- Therefore, for positive roots, \(a \in (-5, 1)\).
2. **Opposite Sign Roots**: For the roots to have opposite signs:
- The product of roots must be negative: \(a + 5 < 0 \Rightarrow a < -5\).
3. **Negative Roots**: For both roots to be negative:
- The sum of roots must be negative: \(-2(a - 1) < 0 \Rightarrow a > 1\)
- The product of roots must be positive: \(a + 5 > 0 \Rightarrow a > -5\)
- Therefore, for negative roots, \(a \in (1, \infty)\).
### Conclusion
Based on the analysis:
- Positive roots: \(a \in (-5, 1)\)
- Opposite sign roots: \(a < -5\)
- Negative roots: \(a \in (1, \infty)\)
### Final Answer
- Option 1: Correct for positive roots \(a \in (-5, -1)\).
- Option 2: Correct for opposite sign roots \(a < -5\).
- Option 3: Incorrect for negative roots (should be \(a > 1\)).
- Option 4: None of these is incorrect.
