To find the value of \( xy \) given the equations
- \( x^2 + y^2 + xy = 1 \)
- \( x + y = 2 \)
we can start by expressing \( y \) in terms of \( x \) using the second equation.
From \( x + y = 2 \), we can isolate \( y \):
\[ y = 2 - x \]
Next, substitute \( y \) into the first equation:
\[ x^2 + (2 - x)^2 + x(2 - x) = 1 \]
Now, expand and simplify each term:
- The term \( (2 - x)^2 \):
\[ (2 - x)^2 = 4 - 4x + x^2 \]
- The term \( x(2 - x) \):
\[ x(2 - x) = 2x - x^2 \]
Now substituting these into the first equation:
\[ x^2 + (4 - 4x + x^2) + (2x - x^2) = 1 \]
Combine the expressions:
\[ x^2 + 4 - 4x + x^2 + 2x - x^2 = 1 \]
Simplifying gives:
\[ x^2 - 2x + 4 = 1 \]
Subtract 1 from both sides:
\[ x^2 - 2x + 3 = 0 \]
Next, we solve the quadratic equation \( x^2 - 2x + 3 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \), and \( c = 3 \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 12}}{2} \] \[ = \frac{2 \pm \sqrt{-8}}{2} = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2} \]
Thus we have:
\[ x = 1 + i\sqrt{2} \quad \text{or} \quad x = 1 - i\sqrt{2} \]
Now substituting back to find \( y \):
- If \( x = 1 + i\sqrt{2} \):
\[ y = 2 - (1 + i\sqrt{2}) = 1 - i\sqrt{2} \]
- If \( x = 1 - i\sqrt{2} \):
\[ y = 2 - (1 - i\sqrt{2}) = 1 + i\sqrt{2} \]
Now we find \( xy \):
\[ xy = (1 + i\sqrt{2})(1 - i\sqrt{2}) = 1^2 - (i\sqrt{2})^2 = 1 - (-2) = 1 + 2 = 3 \]
Thus, the value of \( xy \) is
\[ \boxed{3} \]
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