If one root of `x^(2) - x - k = 0` is square of the other, then k =

Options

A. `2 pm sqrt5`
B. `2 pm sqrt3`
C. `3 pm sqrt2`
D. `5 pm sqrt2`

Explanation

To solve the equation \( x^2 - x - k = 0 \) given that one root is the square of the other, we can follow these steps:

### Step 1: Define the roots
Let the roots of the equation be \( \alpha \) and \( \alpha^2 \), where \( \alpha \) is one root and \( \alpha^2 \) is the square of the other root.

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \alpha^2 = -\frac{b}{a} \)
- The product of the roots \( \alpha \cdot \alpha^2 = \frac{c}{a} \)

For our equation \( x^2 - x - k = 0 \):
- \( a = 1 \), \( b = -1 \), and \( c = -k \)

### Step 3: Set up the equations
From Vieta's formulas, we have:
1. \( \alpha + \alpha^2 = 1 \) (since \( -b/a = 1 \))
2. \( \alpha \cdot \alpha^2 = -k \) (since \( c/a = -k \))

### Step 4: Express \( \alpha^2 \) in terms of \( \alpha \)
From the first equation, we can express \( \alpha^2 \):
\[ \alpha^2 = 1 - \alpha \]

### Step 5: Substitute into the product equation
Substituting \( \alpha^2 \) into the product equation:
\[ \alpha \cdot (1 - \alpha) = -k \]
This simplifies to:
\[ \alpha - \alpha^2 = -k \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ k = \alpha^2 - \alpha \]

### Step 7: Substitute \( \alpha^2 \)
Now substitute \( \alpha^2 = 1 - \alpha \):
\[ k = (1 - \alpha) - \alpha = 1 - 2\alpha \]

### Step 8: Solve for \( \alpha \)
We can substitute \( \alpha^2 = 1 - \alpha \) back into our product equation:
\[ \alpha(1 - \alpha) = -k \]
Substituting \( k = 1 - 2\alpha \):
\[ \alpha(1 - \alpha) = -(1 - 2\alpha) \]
Expanding gives:
\[ \alpha - \alpha^2 = -1 + 2\alpha \]
Rearranging yields:
\[ \alpha^2 - \alpha - 1 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \alpha = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]

### Step 10: Find \( k \)
Now substituting \( \alpha \) back into \( k = 1 - 2\alpha \):
1. For \( \alpha = \frac{1 + \sqrt{5}}{2} \):
\[ k = 1 - 2\left(\frac{1 + \sqrt{5}}{2}\right) = 1 - (1 + \sqrt{5}) = -\sqrt{5} \]
2. For \( \alpha = \frac{1 - \sqrt{5}}{2} \):
\[ k = 1 - 2\left(\frac{1 - \sqrt{5}}{2}\right) = 1 - (1 - \sqrt{5}) = \sqrt{5} \]

### Conclusion
Thus, the possible values for \( k \) are \( k = -\sqrt{5} \) and \( k = \sqrt{5} \).