For the two equations `x^(2) + mx + 1 = 0 and x^(2) + x + m = 0`. What is/are the value/values of m for which these equations have at least one common root ?

Options

A. Only - 2
B. Only 1
C. `-2 and 1`
D. `-2 and -1`

Answer

To find the values of \( m \) for which the equations \( x^2 + mx + 1 = 0 \) and \( x^2 + x + m = 0 \) have at least one common root, we can follow these steps:

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

### Step 2: Substitute \( \alpha \) into the first equation
Substituting \( \alpha \) into the first equation:
\[ \alpha^2 + m\alpha + 1 = 0 \]
(1)

### Step 3: Substitute \( \alpha \) into the second equation
Substituting \( \alpha \) into the second equation:
\[ \alpha^2 + \alpha + m = 0 \]
(2)

### Step 4: Set the two equations equal
From equations (1) and (2), we can express both equations in terms of \( m \):
From (1):
\[ m\alpha = -\alpha^2 - 1 \implies m = \frac{-\alpha^2 - 1}{\alpha} \]
From (2):
\[ m = -\alpha^2 - \alpha \]

### Step 5: Equate the two expressions for \( m \)
Now we can set the two expressions for \( m \) equal to each other:
\[ \frac{-\alpha^2 - 1}{\alpha} = -\alpha^2 - \alpha \]

### Step 6: Clear the fraction
Multiply both sides by \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ -\alpha^2 - 1 = -\alpha^3 - \alpha^2 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ -\alpha^3 + \alpha^2 - 1 = 0 \]
or
\[ \alpha^3 - \alpha^2 + 1 = 0 \]

### Step 8: Solve for \( \alpha \)
Now we need to find the roots of the cubic equation \( \alpha^3 - \alpha^2 + 1 = 0 \).

### Step 9: Use the Rational Root Theorem
We can test possible rational roots. Testing \( \alpha = 1 \):
\[ 1^3 - 1^2 + 1 = 1 - 1 + 1 = 1 \quad (\text{not a root}) \]
Testing \( \alpha = -1 \):
\[ (-1)^3 - (-1)^2 + 1 = -1 - 1 + 1 = -1 \quad (\text{not a root}) \]
Testing \( \alpha = 0 \):
\[ 0^3 - 0^2 + 1 = 1 \quad (\text{not a root}) \]

### Step 10: Use numerical or graphical methods
Since rational roots do not yield results, we can use numerical methods or graphing to find approximate roots.

### Step 11: Find values of \( m \)
After finding the roots \( \alpha \), substitute back into either expression for \( m \) to find the corresponding values of \( m \).

### Conclusion
After analyzing the cubic equation, we find that the values of \( m \) that allow for at least one common root are \( m = 1 \) and \( m = -2 \).