If `x^(2)+y^(2)=156` and `x-y=12`, then `xy=?`

Options

A. 8
B. 13
C. 11
D. 6

Answer

To solve the problem, we are given two equations:

1. \( x^2 + y^2 = 156 \)
2. \( x - y = 12 \)

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

### Step 1: Express \( x \) in terms of \( y \)
From the second equation, we can express \( x \) in terms of \( y \):
\[ x = y + 12 \]

### Step 2: Substitute \( x \) in the first equation
Now, we substitute \( x \) in the first equation:
\[ (y + 12)^2 + y^2 = 156 \]

### Step 3: Expand the equation
Expanding the left-hand side:
\[ (y^2 + 24y + 144) + y^2 = 156 \]

### Step 4: Combine like terms
Combine the \( y^2 \) terms:
\[ 2y^2 + 24y + 144 = 156 \]

### Step 5: Rearrange the equation
Now, we rearrange the equation to set it to zero:
\[ 2y^2 + 24y + 144 - 156 = 0 \]
\[ 2y^2 + 24y - 12 = 0 \]

### Step 6: Simplify the equation
We can simplify the equation by dividing all terms by 2:
\[ y^2 + 12y - 6 = 0 \]

### Step 7: Use the quadratic formula
Now, we can use the quadratic formula to find \( y \):
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 12 \), and \( c = -6 \):
\[ y = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ y = \frac{-12 \pm \sqrt{144 + 24}}{2} \]
\[ y = \frac{-12 \pm \sqrt{168}}{2} \]
\[ y = \frac{-12 \pm 2\sqrt{42}}{2} \]
\[ y = -6 \pm \sqrt{42} \]

### Step 8: Find \( x \)
Now we can find \( x \):
\[ x = y + 12 = (-6 \pm \sqrt{42}) + 12 = 6 \pm \sqrt{42} \]

### Step 9: Calculate \( xy \)
Now we can find \( xy \):
\[ xy = x \cdot y = (6 + \sqrt{42})(-6 + \sqrt{42}) \text{ or } (6 - \sqrt{42})(-6 - \sqrt{42}) \]
Using the difference of squares:
\[ xy = (6^2 - (\sqrt{42})^2) = 36 - 42 = -6 \]

### Final Answer
Thus, the value of \( xy \) is:
\[ \boxed{-6} \]