if `x^3+ax+1=0 `and `x^4+ax^2+1=0` have common root then the exhaustive set of value of `a` is

Options

A. `a=2`
B. `a=-2`
C. `a=0`
D. none of these

Explanation

To solve the problem, we need to find the values of \( a \) such that the equations \( x^3 + ax + 1 = 0 \) and \( x^4 + ax^2 + 1 = 0 \) have a common root.

### Step-by-Step Solution:

1. **Let the common root be \( \alpha \)**:
Since both equations share a common root, we can denote this root as \( \alpha \). Therefore, we can write:
\[ \alpha^3 + a\alpha + 1 = 0 \quad \text{(Equation 1)} \]
\[ \alpha^4 + a\alpha^2 + 1 = 0 \quad \text{(Equation 2)} \]

2. **Subtract Equation 1 from Equation 2**:
We can eliminate \( a \) by subtracting Equation 1 from Equation 2:
\[ (\alpha^4 + a\alpha^2 + 1) - (\alpha^3 + a\alpha + 1) = 0 \]
Simplifying this gives:
\[ \alpha^4 - \alpha^3 + a\alpha^2 - a\alpha = 0 \]

3. **Factor the resulting equation**:
We can factor out \( \alpha^3 \) and \( a\alpha \):
\[ \alpha^3(\alpha - 1) + a\alpha(\alpha - 1) = 0 \]
Factoring out \( \alpha - 1 \):
\[ (\alpha - 1)(\alpha^3 + a\alpha) = 0 \]

4. **Set the factors to zero**:
This gives us two cases:
- Case 1: \( \alpha - 1 = 0 \) → \( \alpha = 1 \)
- Case 2: \( \alpha^3 + a\alpha = 0 \)

5. **Solve for \( a \) in Case 1**:
If \( \alpha = 1 \):
Substitute \( \alpha = 1 \) into Equation 1:
\[ 1^3 + a(1) + 1 = 0 \implies 1 + a + 1 = 0 \implies a + 2 = 0 \implies a = -2 \]

6. **Solve for \( a \) in Case 2**:
If \( \alpha^3 + a\alpha = 0 \):
Rearranging gives:
\[ a = -\frac{\alpha^3}{\alpha} = -\alpha^2 \quad (\text{for } \alpha \neq 0) \]

7. **Substituting \( a = -\alpha^2 \) into Equation 1**:
Substitute \( a = -\alpha^2 \) into Equation 1:
\[ \alpha^3 - \alpha^2\alpha + 1 = 0 \implies \alpha^3 - \alpha^3 + 1 = 0 \implies 1 = 0 \quad \text{(not possible)} \]

8. **Check the case \( \alpha = 0 \)**:
If \( \alpha = 0 \):
Substitute into Equation 1:
\[ 0 + a(0) + 1 = 0 \implies 1 = 0 \quad \text{(not valid)} \]

### Conclusion:
The only valid solution for \( a \) is when \( \alpha = 1 \), leading to:
\[ \text{The exhaustive set of values of } a \text{ is } \{ -2 \}. \]