Are the following statements True or False Justify your answers.
The only value of k for which the quadratic polynomial `kx^(2)+x+k` has equal zeroes is `(1)/(2)` .

Answer

To determine whether the statement "The only value of k for which the quadratic polynomial \( kx^2 + x + k \) has equal zeroes is \( \frac{1}{2} \)" is true or false, we can follow these steps:

### Step 1: Identify the quadratic polynomial
The given quadratic polynomial is:
\[ kx^2 + x + k \]
where \( a = k \), \( b = 1 \), and \( c = k \).

### Step 2: Use the condition for equal roots
For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant must be zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = 1^2 - 4(k)(k) = 1 - 4k^2 \]

### Step 3: Set the discriminant to zero
To find the values of \( k \) for which the roots are equal, we set the discriminant to zero:
\[ 1 - 4k^2 = 0 \]

### Step 4: Solve for \( k \)
Rearranging the equation gives:
\[ 4k^2 = 1 \]
Dividing both sides by 4:
\[ k^2 = \frac{1}{4} \]
Taking the square root of both sides, we get:
\[ k = \frac{1}{2} \quad \text{or} \quad k = -\frac{1}{2} \]

### Step 5: Conclusion
The values of \( k \) for which the polynomial has equal roots are \( \frac{1}{2} \) and \( -\frac{1}{2} \). Since the statement claims that the only value is \( \frac{1}{2} \), it is incorrect.

### Final Answer
The statement is **False** because there are two values of \( k \) (i.e., \( \frac{1}{2} \) and \( -\frac{1}{2} \)) for which the polynomial has equal zeroes.

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