If the equations `x^(2)+ax+12=0,x^(2)+bx+15=0` and `x^(2)+(a+b)x+36=0` have a common positive root, then the ordered pair (a, b) is

Options

A. `(-6,-8)`
B. `(-7,-8)`
C. `(-6,-7)`
D. `(-7,-9)`

Answer

To solve the problem, we need to find the ordered pair (a, b) such that the equations \(x^2 + ax + 12 = 0\), \(x^2 + bx + 15 = 0\), and \(x^2 + (a+b)x + 36 = 0\) have a common positive root.

### Step-by-Step Solution:

1. **Let the common positive root be \( \alpha \)**.
- For the first equation \(x^2 + ax + 12 = 0\):
- By Vieta's formulas, we have:
\[ \alpha + \beta = -a \quad \text{(1)} \]
\[ \alpha \beta = 12 \quad \text{(2)} \]

2. **For the second equation \(x^2 + bx + 15 = 0\)**:
- Again using Vieta's formulas:
- Let the other root be \( \gamma \):
\[ \alpha + \gamma = -b \quad \text{(3)} \]
\[ \alpha \gamma = 15 \quad \text{(4)} \]

3. **For the third equation \(x^2 + (a+b)x + 36 = 0\)**:
- Let the other root be \( \delta \):
\[ \alpha + \delta = -(a+b) \quad \text{(5)} \]
\[ \alpha \delta = 36 \quad \text{(6)} \]

4. **Express \( \beta \), \( \gamma \), and \( \delta \) in terms of \( \alpha \)**:
- From (2):
\[ \beta = \frac{12}{\alpha} \]
- From (4):
\[ \gamma = \frac{15}{\alpha} \]
- From (6):
\[ \delta = \frac{36}{\alpha} \]

5. **Substituting values into (1), (3), and (5)**:
- Substitute \( \beta \) into (1):
\[ \alpha + \frac{12}{\alpha} = -a \quad \Rightarrow \quad a = -\left(\alpha + \frac{12}{\alpha}\right) \quad \text{(7)} \]
- Substitute \( \gamma \) into (3):
\[ \alpha + \frac{15}{\alpha} = -b \quad \Rightarrow \quad b = -\left(\alpha + \frac{15}{\alpha}\right) \quad \text{(8)} \]
- Substitute \( \delta \) into (5):
\[ \alpha + \frac{36}{\alpha} = -(a+b) \quad \Rightarrow \quad a + b = -\left(\alpha + \frac{36}{\alpha}\right) \quad \text{(9)} \]

6. **Using (7) and (8) in (9)**:
- Substitute \( a \) and \( b \) from (7) and (8):
\[ -\left(\alpha + \frac{12}{\alpha}\right) - \left(\alpha + \frac{15}{\alpha}\right) = -\left(\alpha + \frac{36}{\alpha}\right) \]
- Simplifying gives:
\[ -2\alpha - \frac{27}{\alpha} = -\alpha - \frac{36}{\alpha} \]
- Rearranging:
\[ -\alpha = -\frac{9}{\alpha} \]
- Multiplying both sides by \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ \alpha^2 = 9 \quad \Rightarrow \quad \alpha = 3 \quad \text{(since we need the positive root)} \]

7. **Finding \( a \) and \( b \)**:
- Substitute \( \alpha = 3 \) back into (7) and (8):
\[ a = -\left(3 + \frac{12}{3}\right) = -\left(3 + 4\right) = -7 \]
\[ b = -\left(3 + \frac{15}{3}\right) = -\left(3 + 5\right) = -8 \]

8. **Final Ordered Pair**:
- The ordered pair \( (a, b) \) is:
\[ (a, b) = (-7, -8) \]