If `x^2-ax-45=0` and `x^2-3ax+27=0,(agt0)` has a has a common root then the value of a is

Options

A. a) 5
B. b) `pm7`
C. c) `pm8`
D. d) 4

Answer

To solve the problem, we need to find the value of \( a \) such that the equations \( x^2 - ax - 45 = 0 \) and \( x^2 - 3ax + 27 = 0 \) have a common root. Let's denote the common root as \( p \).

### Step-by-Step Solution:

1. **Substituting the Common Root in Both Equations:**
- For the first equation \( x^2 - ax - 45 = 0 \):
\[ p^2 - ap - 45 = 0 \quad \text{(1)} \]
- For the second equation \( x^2 - 3ax + 27 = 0 \):
\[ p^2 - 3ap + 27 = 0 \quad \text{(2)} \]

2. **Equating the Two Equations:**
Since both equations equal zero at the common root \( p \), we can set the two equations equal to each other:
\[ p^2 - ap - 45 = p^2 - 3ap + 27 \]

3. **Simplifying the Equation:**
- Cancel \( p^2 \) from both sides:
\[ -ap - 45 = -3ap + 27 \]
- Rearranging gives:
\[ -ap + 3ap = 27 + 45 \]
\[ 2ap = 72 \]

4. **Solving for \( a \):**
- Divide both sides by \( 2p \) (assuming \( p \neq 0 \)):
\[ a = \frac{72}{2p} = \frac{36}{p} \quad \text{(3)} \]

5. **Substituting \( a \) Back into One of the Original Equations:**
- We can substitute \( a \) from equation (3) back into the first equation (1):
\[ p^2 - \left(\frac{36}{p}\right)p - 45 = 0 \]
- This simplifies to:
\[ p^2 - 36 - 45 = 0 \]
\[ p^2 - 81 = 0 \]

6. **Finding the Value of \( p \):**
- Solving for \( p \):
\[ p^2 = 81 \]
\[ p = 9 \quad \text{(since \( p > 0 \))} \]

7. **Finding the Final Value of \( a \):**
- Substitute \( p = 9 \) back into equation (3):
\[ a = \frac{36}{9} = 4 \]

### Final Answer:
The value of \( a \) is \( 4 \).