Can a quadratic polynomial `x^2 + kx + k` have equal zeros for some odd integer `k gt 1`?

Answer

To determine whether the quadratic polynomial \( x^2 + kx + k \) can have equal zeros for some odd integer \( k > 1 \), we can follow these steps:

### Step 1: Identify the conditions for equal zeros
For a quadratic polynomial of the form \( ax^2 + bx + c \), the zeros are equal if the discriminant \( D \) is zero. The discriminant is given by:
\[ D = b^2 - 4ac \]

### Step 2: Substitute the values into the discriminant
In our case, the polynomial is \( x^2 + kx + k \), where:
- \( a = 1 \)
- \( b = k \)
- \( c = k \)

Thus, the discriminant becomes:
\[ D = k^2 - 4(1)(k) = k^2 - 4k \]

### Step 3: Set the discriminant to zero for equal zeros
To find the condition for equal zeros, we set the discriminant equal to zero:
\[ k^2 - 4k = 0 \]

### Step 4: Factor the equation
Factoring the equation gives:
\[ k(k - 4) = 0 \]

### Step 5: Solve for \( k \)
Setting each factor to zero gives us:
1. \( k = 0 \)
2. \( k - 4 = 0 \) which gives \( k = 4 \)

### Step 6: Analyze the solutions
The solutions we found are \( k = 0 \) and \( k = 4 \). However, we need to check if these values satisfy the condition that \( k \) is an odd integer greater than 1.

- \( k = 0 \) is not greater than 1.
- \( k = 4 \) is an even integer.

### Conclusion
Since both solutions are either not odd or not greater than 1, we conclude that there is no odd integer \( k > 1 \) for which the quadratic polynomial \( x^2 + kx + k \) has equal zeros.