If `(x+y)^(2)-(x-y)^(2)=28`, what is the value of xy?

Answer

To solve the equation \((x+y)^{2} - (x-y)^{2} = 28\) and find the value of \(xy\), we can follow these steps:

### Step 1: Expand both squares
Using the identity for the square of a binomial, we can expand both \((x+y)^2\) and \((x-y)^2\):

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

### Step 2: Substitute the expansions into the equation
Now, substitute these expansions back into the original equation:

\[ (x^{2} + 2xy + y^{2}) - (x^{2} - 2xy + y^{2}) = 28 \]

### Step 3: Simplify the equation
Now simplify the left-hand side:

\[ x^{2} + 2xy + y^{2} - x^{2} + 2xy - y^{2} = 28 \]

The \(x^{2}\) and \(y^{2}\) terms cancel out:

\[ 4xy = 28 \]

### Step 4: Solve for \(xy\)
Now, divide both sides by 4 to isolate \(xy\):

\[ xy = \frac{28}{4} = 7 \]

### Final Answer
Thus, the value of \(xy\) is:

\[ \boxed{7} \]

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