If the equation `x^(2) + k^(2) = 2 (k+1)x` has equal roots, then what is the value of K?

Options

A. `-1/3`
B. `-1/2`
C. 0
D. 1

Explanation

To solve the equation \(x^2 + k^2 = 2(k + 1)x\) for the value of \(k\) such that the equation has equal roots, we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation into standard quadratic form:
\[ x^2 - 2(k + 1)x + k^2 = 0 \]
Here, we identify the coefficients:
- \(a = 1\)
- \(b = -2(k + 1)\)
- \(c = k^2\)

### Step 2: Condition for Equal Roots
For a quadratic equation to have equal roots, the discriminant must be zero. The discriminant \(D\) is given by:
\[ D = b^2 - 4ac \]
Setting the discriminant to zero gives us:
\[ 0 = (-2(k + 1))^2 - 4(1)(k^2) \]

### Step 3: Expanding the Discriminant
Now, we expand the discriminant:
\[ 0 = 4(k + 1)^2 - 4k^2 \]
\[ 0 = 4(k^2 + 2k + 1) - 4k^2 \]
\[ 0 = 4k^2 + 8k + 4 - 4k^2 \]
\[ 0 = 8k + 4 \]

### Step 4: Solving for \(k\)
Now, we can solve for \(k\):
\[ 8k + 4 = 0 \]
\[ 8k = -4 \]
\[ k = -\frac{4}{8} = -\frac{1}{2} \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{-\frac{1}{2}} \]