If `x+y=2k-1, and x^(2)+y^(2)=9-4k+2k^(2)`, what is xyin terms of k?

Options

A. `k-2`
B. `(k-2)^(2)`
C. `(k+2)^(2)`
D. `k^(2)-4`

Answer

To find \( xy \) in terms of \( k \), we start with the two equations given:

1. \( x + y = 2k - 1 \)
2. \( x^2 + y^2 = 9 - 4k + 2k^2 \)

We will use the identity for the square of a sum:

\[ (x + y)^2 = x^2 + y^2 + 2xy \]

### Step 1: Substitute the values into the identity

Substituting the values from the equations into the identity:

\[ (2k - 1)^2 = x^2 + y^2 + 2xy \]

### Step 2: Expand the left-hand side

Now, we expand the left-hand side:

\[ (2k - 1)^2 = 4k^2 - 4k + 1 \]

So we have:

\[ 4k^2 - 4k + 1 = x^2 + y^2 + 2xy \]

### Step 3: Substitute \( x^2 + y^2 \)

Now we substitute \( x^2 + y^2 \) from the second equation:

\[ 4k^2 - 4k + 1 = (9 - 4k + 2k^2) + 2xy \]

### Step 4: Rearrange the equation

Rearranging gives us:

\[ 4k^2 - 4k + 1 - 9 + 4k - 2k^2 = 2xy \]

### Step 5: Simplify the equation

Now we simplify:

\[ (4k^2 - 2k^2) + (-4k + 4k) + (1 - 9) = 2xy \]

This simplifies to:

\[ 2k^2 - 8 = 2xy \]

### Step 6: Solve for \( xy \)

Now, we divide both sides by 2:

\[ k^2 - 4 = xy \]

Thus, we have:

\[ xy = k^2 - 4 \]

### Final Answer

The value of \( xy \) in terms of \( k \) is:

\[ \boxed{k^2 - 4} \]