If `x^4 + y^4 + x^2 y^2 = 117 and x^2 + y^2 + xy = 13` then find the value of `x^2 + y^2`.

Options

A. 9
B. 13
C. 7
D. 11

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 \) given the equations:

1. \( x^4 + y^4 + x^2y^2 = 117 \)
2. \( x^2 + y^2 + xy = 13 \)

Let's denote \( a = x^2 + y^2 \) and \( b = xy \).

### Step 1: Rewrite the first equation
We can express \( x^4 + y^4 \) in terms of \( a \) and \( b \):
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = a^2 - 2b^2 \]
Thus, the first equation becomes:
\[ a^2 - 2b^2 + b^2 = 117 \]
This simplifies to:
\[ a^2 - b^2 = 117 \]

### Step 2: Rewrite the second equation
The second equation is already in terms of \( a \) and \( b \):
\[ a + b = 13 \]
From this, we can express \( b \) in terms of \( a \):
\[ b = 13 - a \]

### Step 3: Substitute \( b \) into the first equation
Substituting \( b = 13 - a \) into the equation \( a^2 - b^2 = 117 \):
\[ a^2 - (13 - a)^2 = 117 \]
Expanding \( (13 - a)^2 \):
\[ a^2 - (169 - 26a + a^2) = 117 \]
This simplifies to:
\[ a^2 - 169 + 26a - a^2 = 117 \]
Thus:
\[ 26a - 169 = 117 \]

### Step 4: Solve for \( a \)
Rearranging gives:
\[ 26a = 117 + 169 \]
\[ 26a = 286 \]
Dividing both sides by 26:
\[ a = \frac{286}{26} = 11 \]

### Conclusion
The value of \( x^2 + y^2 \) is \( 11 \).