If x = 1 - y and `x^(2) = 2 - y^(2)` , what is the value of xy ?

Options

A. 1
B. 2
C. `(-1)/(2)`
D. `-1`

Answer

To solve the problem, we start with the given equations:

1. \( x = 1 - y \)
2. \( x^2 = 2 - y^2 \)

We need to find the value of \( xy \).

### Step 1: Substitute \( x \) in the second equation

From the first equation, we can express \( x \) in terms of \( y \):
\[ x = 1 - y \]

Now, substitute this expression for \( x \) into the second equation:
\[ (1 - y)^2 = 2 - y^2 \]

### Step 2: Expand the left side of the equation

Expanding \( (1 - y)^2 \):
\[ 1 - 2y + y^2 = 2 - y^2 \]

### Step 3: Rearrange the equation

Now, rearranging the equation gives:
\[ 1 - 2y + y^2 + y^2 - 2 = 0 \]
\[ 2y^2 - 2y - 1 = 0 \]

### Step 4: Simplify the equation

Dividing the entire equation by 2:
\[ y^2 - y - \frac{1}{2} = 0 \]

### Step 5: Solve the quadratic equation

We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = -\frac{1}{2} \):
\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot -\frac{1}{2}}}{2 \cdot 1} \]
\[ y = \frac{1 \pm \sqrt{1 + 2}}{2} \]
\[ y = \frac{1 \pm \sqrt{3}}{2} \]

### Step 6: Find the corresponding values of \( x \)

Using the values of \( y \) to find \( x \):
1. For \( y = \frac{1 + \sqrt{3}}{2} \):
\[ x = 1 - y = 1 - \frac{1 + \sqrt{3}}{2} = \frac{1 - \sqrt{3}}{2} \]

2. For \( y = \frac{1 - \sqrt{3}}{2} \):
\[ x = 1 - y = 1 - \frac{1 - \sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \]

### Step 7: Calculate \( xy \)

Now we calculate \( xy \) for both pairs of \( (x, y) \):
1. For \( x = \frac{1 - \sqrt{3}}{2} \) and \( y = \frac{1 + \sqrt{3}}{2} \):
\[ xy = \left(\frac{1 - \sqrt{3}}{2}\right) \left(\frac{1 + \sqrt{3}}{2}\right) = \frac{(1 - \sqrt{3})(1 + \sqrt{3})}{4} = \frac{1 - 3}{4} = \frac{-2}{4} = -\frac{1}{2} \]

2. For \( x = \frac{1 + \sqrt{3}}{2} \) and \( y = \frac{1 - \sqrt{3}}{2} \):
\[ xy = \left(\frac{1 + \sqrt{3}}{2}\right) \left(\frac{1 - \sqrt{3}}{2}\right) = \frac{(1 + \sqrt{3})(1 - \sqrt{3})}{4} = \frac{1 - 3}{4} = -\frac{1}{2} \]

### Final Answer

Thus, in both cases, the value of \( xy \) is:
\[ \boxed{-\frac{1}{2}} \]