If `x^(2) + y^(2) = 47` and `xy = (19)/(2)`, then the value of `2(x + y)^(2) + (x-y)^(2)` is

Options

A. 160
B. 270
C. 226
D. 86

Answer

To solve the problem, we need to find the value of \(2(x + y)^2 + (x - y)^2\) given that \(x^2 + y^2 = 47\) and \(xy = \frac{19}{2}\).

### Step-by-step Solution:

1. **Expand the expression**:
\[ 2(x + y)^2 + (x - y)^2 \]
We can expand these squares:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
\[ (x - y)^2 = x^2 + y^2 - 2xy \]

2. **Substitute the expansions into the expression**:
\[ 2(x + y)^2 + (x - y)^2 = 2(x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) \]

3. **Combine like terms**:
\[ = 2(x^2 + y^2) + 4xy + (x^2 + y^2) - 2xy \]
\[ = (2x^2 + 2y^2 + 4xy + x^2 + y^2 - 2xy) \]
\[ = (3x^2 + 3y^2 + 2xy) \]

4. **Factor out the common terms**:
\[ = 3(x^2 + y^2) + 2xy \]

5. **Substitute the known values**:
We know \(x^2 + y^2 = 47\) and \(xy = \frac{19}{2}\):
\[ = 3(47) + 2\left(\frac{19}{2}\right) \]
\[ = 141 + 19 \]

6. **Calculate the final result**:
\[ = 160 \]

Thus, the value of \(2(x + y)^2 + (x - y)^2\) is **160**.