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

Options

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

Explanation

To find the value 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. Therefore, we can write:
1. \( \alpha^2 + m\alpha + 1 = 0 \) (Equation 1)
2. \( \alpha^2 + \alpha + m = 0 \) (Equation 2)

### Step 2: Set the equations equal
Since both equations equal zero, we can set them equal to each other:
\[ \alpha^2 + m\alpha + 1 = \alpha^2 + \alpha + m \]

### Step 3: Simplify the equation
Subtract \( \alpha^2 \) from both sides:
\[ m\alpha + 1 = \alpha + m \]
Rearranging gives us:
\[ m\alpha - \alpha + 1 - m = 0 \]
Factoring out \( \alpha \):
\[ \alpha(m - 1) + (1 - m) = 0 \]

### Step 4: Factor the equation
This can be factored as:
\[ (m - 1)(\alpha - 1) = 0 \]

### Step 5: Solve for \( m \)
This gives us two cases:
1. \( m - 1 = 0 \) which implies \( m = 1 \)
2. \( \alpha - 1 = 0 \) which implies \( \alpha = 1 \)

### Step 6: Substitute \( \alpha = 1 \) back into either equation
Substituting \( \alpha = 1 \) into Equation 1:
\[ 1^2 + m(1) + 1 = 0 \implies 1 + m + 1 = 0 \implies m + 2 = 0 \implies m = -2 \]

### Step 7: Conclusion
Thus, the values of \( m \) for which the equations have at least one common root are:
\[ m = 1 \quad \text{and} \quad m = -2 \]

### Final Answer
The values of \( m \) are \( 1 \) and \( -2 \).

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