The value of a such that `x^2-11 x+a=0` and `x^2-14 x+2a=0` may have a common root is

Options

A. `0` & `24`
B. `0` & `12`
C. `2` & `24`
D. `12` & `32`

Answer

To find the value of \( a \) such that the equations \( x^2 - 11x + a = 0 \) and \( x^2 - 14x + 2a = 0 \) have a common root, we can follow these steps:

### Step 1: Assume a Common Root
Let the common root be \( \alpha \). This means that \( \alpha \) satisfies both equations.

### Step 2: Set Up the Equations
Substituting \( \alpha \) into the first equation:
\[ \alpha^2 - 11\alpha + a = 0 \quad \text{(1)} \]
Substituting \( \alpha \) into the second equation:
\[ \alpha^2 - 14\alpha + 2a = 0 \quad \text{(2)} \]

### Step 3: Subtract the Equations
Now, we can subtract equation (1) from equation (2):
\[ (\alpha^2 - 14\alpha + 2a) - (\alpha^2 - 11\alpha + a) = 0 \]
This simplifies to:
\[ -14\alpha + 2a + 11\alpha - a = 0 \]
Combining like terms gives:
\[ -3\alpha + a = 0 \]
Thus, we can express \( a \) in terms of \( \alpha \):
\[ a = 3\alpha \quad \text{(3)} \]

### Step 4: Substitute \( a \) Back
Now, substitute equation (3) back into either of the original equations. We will use the first equation:
\[ \alpha^2 - 11\alpha + 3\alpha = 0 \]
This simplifies to:
\[ \alpha^2 - 8\alpha = 0 \]

### Step 5: Factor the Equation
Factoring out \( \alpha \):
\[ \alpha(\alpha - 8) = 0 \]
This gives us two possible values for \( \alpha \):
\[ \alpha = 0 \quad \text{or} \quad \alpha = 8 \]

### Step 6: Find Corresponding Values of \( a \)
Using \( a = 3\alpha \):
1. If \( \alpha = 0 \):
\[ a = 3 \times 0 = 0 \]
2. If \( \alpha = 8 \):
\[ a = 3 \times 8 = 24 \]

### Conclusion
The possible values of \( a \) are \( 0 \) and \( 24 \).

### Final Answer
Thus, the values of \( a \) such that the two equations have a common root are \( a = 0 \) and \( a = 24 \).

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