If the equation `x ^(2) +ax+12 =9, x ^(2) +bx +15 =0 and x^(2) + (a+b) x +36=0` have a common positive root, then `b+2a` equal to.

Options

A. `-6`
B. 22
C. 6
D. `-22`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript while ensuring clarity and organization.

### Step 1: Define the Common Root
Let the common positive root of the equations be denoted as \( \alpha \).

### Step 2: Write the Equations for Roots
From the equations given:
1. \( x^2 + ax + 12 = 9 \) can be rewritten as \( x^2 + ax + 3 = 0 \).
- The sum of the roots \( \alpha + \beta_1 = -a \) (where \( \beta_1 \) is the other root).
- The product of the roots \( \alpha \beta_1 = 3 \).

2. \( x^2 + bx + 15 = 0 \).
- The sum of the roots \( \alpha + \beta_2 = -b \) (where \( \beta_2 \) is the other root).
- The product of the roots \( \alpha \beta_2 = 15 \).

3. \( x^2 + (a+b)x + 36 = 0 \).
- The sum of the roots \( \alpha + \beta_3 = -(a+b) \) (where \( \beta_3 \) is the other root).
- The product of the roots \( \alpha \beta_3 = 36 \).

### Step 3: Set Up the Relationships
From the equations, we have:
1. \( \alpha + \beta_1 = -a \) (1)
2. \( \alpha \beta_1 = 3 \) (2)
3. \( \alpha + \beta_2 = -b \) (3)
4. \( \alpha \beta_2 = 15 \) (4)
5. \( \alpha + \beta_3 = -(a+b) \) (5)
6. \( \alpha \beta_3 = 36 \) (6)

### Step 4: Express Other Roots in Terms of \( \alpha \)
From (2):
\[ \beta_1 = \frac{3}{\alpha} \]
From (4):
\[ \beta_2 = \frac{15}{\alpha} \]
From (6):
\[ \beta_3 = \frac{36}{\alpha} \]

### Step 5: Substitute and Simplify
Substituting \( \beta_1 \) and \( \beta_2 \) into (1) and (3):
1. From (1):
\[ \alpha + \frac{3}{\alpha} = -a \implies a = -\left(\alpha + \frac{3}{\alpha}\right) \]
2. From (3):
\[ \alpha + \frac{15}{\alpha} = -b \implies b = -\left(\alpha + \frac{15}{\alpha}\right) \]

### Step 6: Substitute into (5)
Substituting \( a \) and \( b \) into (5):
\[ \alpha + \frac{36}{\alpha} = -\left(-\left(\alpha + \frac{3}{\alpha}\right) + -\left(\alpha + \frac{15}{\alpha}\right)\right) \]
This simplifies to:
\[ \alpha + \frac{36}{\alpha} = \left(\alpha + \frac{3}{\alpha}\right) + \left(\alpha + \frac{15}{\alpha}\right) \]
\[ \alpha + \frac{36}{\alpha} = 2\alpha + \frac{18}{\alpha} \]
Rearranging gives:
\[ \frac{36}{\alpha} - \frac{18}{\alpha} = 2\alpha - \alpha \]
\[ \frac{18}{\alpha} = \alpha \]
Multiplying both sides by \( \alpha \):
\[ 18 = \alpha^2 \implies \alpha = 3 \quad (\text{since } \alpha \text{ is positive}) \]

### Step 7: Find \( a \) and \( b \)
Substituting \( \alpha = 3 \):
1. For \( a \):
\[ a = -\left(3 + \frac{3}{3}\right) = -\left(3 + 1\right) = -4 \]
2. For \( b \):
\[ b = -\left(3 + \frac{15}{3}\right) = -\left(3 + 5\right) = -8 \]

### Step 8: Calculate \( b + 2a \)
Now we calculate \( b + 2a \):
\[ b + 2a = -8 + 2(-4) = -8 - 8 = -16 \]

### Final Answer
Thus, the value of \( b + 2a \) is:
\[ \boxed{-16} \]