The equations `x^(2)+x+a=0` and `x^(2)+ax+1=0` have a common real root
(a) for no value of a.
(b) for exactly one value of a.
(c) for exactly two values of a.
(d) for exactly three values of a.

Answer

To solve the problem, we need to find the values of \( a \) such that the equations \( x^2 + x + a = 0 \) and \( x^2 + ax + 1 = 0 \) have a common real root. Let's denote the common root as \( \alpha \).

### Step 1: Set up the equations with the common root
We substitute \( \alpha \) into both equations:
1. From the first equation:
\[ \alpha^2 + \alpha + a = 0 \tag{1} \]
2. From the second equation:
\[ \alpha^2 + a\alpha + 1 = 0 \tag{2} \]

### Step 2: Rearranging the equations
From equation (1), we can express \( a \):
\[ a = -(\alpha^2 + \alpha) \tag{3} \]

From equation (2), we can also express \( a \):
\[ a = -\frac{\alpha^2 - 1}{\alpha} \tag{4} \]

### Step 3: Equate the two expressions for \( a \)
Now we set the two expressions for \( a \) equal to each other:
\[ -(\alpha^2 + \alpha) = -\frac{\alpha^2 - 1}{\alpha} \]

### Step 4: Clear the negative sign and multiply through by \( \alpha \)
Multiplying through by \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ \alpha(\alpha^2 + \alpha) = \alpha^2 - 1 \]
This simplifies to:
\[ \alpha^3 + \alpha^2 = \alpha^2 - 1 \]
\[ \alpha^3 = -1 \]

### Step 5: Solve for \( \alpha \)
The solution to \( \alpha^3 + 1 = 0 \) gives:
\[ \alpha^3 = -1 \implies \alpha = -1 \]

### Step 6: Substitute \( \alpha = -1 \) back to find \( a \)
Substituting \( \alpha = -1 \) into equation (3) to find \( a \):
\[ a = -((-1)^2 + (-1)) = -1 + 1 = 0 \]

### Step 7: Check if \( a = 0 \) is valid
Now we check if \( a = 0 \) gives us a common root:
1. For \( a = 0 \):
- The first equation becomes \( x^2 + x = 0 \) which factors to \( x(x + 1) = 0 \) giving roots \( x = 0 \) and \( x = -1 \).
- The second equation becomes \( x^2 + 1 = 0 \) which has no real roots.

Thus, \( a = 0 \) does not provide a common real root.

### Step 8: Analyze the conditions for \( a \)
Now we need to analyze the discriminants of both equations to find conditions for real roots:
1. For \( x^2 + x + a = 0 \):
\[ D_1 = 1 - 4a \geq 0 \implies a \leq \frac{1}{4} \]
2. For \( x^2 + ax + 1 = 0 \):
\[ D_2 = a^2 - 4 \geq 0 \implies a \leq -2 \text{ or } a \geq 2 \]

### Step 9: Combine conditions
The only overlap for real roots occurs when:
- \( a \leq -2 \) or \( a \geq 2 \) and \( a \leq \frac{1}{4} \)

Thus, there are no values of \( a \) for which both equations have a common real root.

### Conclusion
The answer is:
**(a) for no value of a.**