Question

1f x^2 + y^2 + xy = 1 and x + y = 2, then find the value of xy.

Answer 1

To find the value of \( xy \) given the equations

1. \( x^2 + y^2 + xy = 1 \)
2. \( x + y = 2 \)

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:

1. The term \( (2 - x)^2 \):

\[
(2 - x)^2 = 4 - 4x + x^2
\]

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 \( y \):

1. If \( x = 1 + i\sqrt{2} \):

\[
y = 2 - (1 + i\sqrt{2}) = 1 - i\sqrt{2}
\]

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

\[
\boxed{3}
\]