If sum of squares of the roots of the equation `x^(2) + kx - b= 0` is 2b, what is k equal to ?

Options

A. 1
B. b
C. `-b`
D. 0

Explanation

To solve the problem, we need to find the value of \( k \) given that the sum of the squares of the roots of the quadratic equation \( x^2 + kx - b = 0 \) is equal to \( 2b \).

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the equation \( x^2 + kx - b = 0 \) be \( \alpha \) and \( \beta \).

2. **Sum of the Roots**:
From Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{k}{1} = -k \]
and
\[ \alpha \beta = \frac{-b}{1} = -b \]

3. **Sum of the Squares of the Roots**:
The sum of the squares of the roots can be expressed as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (-k)^2 - 2(-b) = k^2 + 2b \]

4. **Set Up the Equation**:
According to the problem, the sum of the squares of the roots is equal to \( 2b \):
\[ k^2 + 2b = 2b \]

5. **Simplify the Equation**:
Subtract \( 2b \) from both sides:
\[ k^2 = 0 \]

6. **Solve for \( k \)**:
Taking the square root of both sides gives:
\[ k = 0 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{0} \).