If `16qxy = (2x + y) ^(2) - (2x - y) ^(2),` then what is the value of q ?

Options

A. `0.5`
B. `0.25`
C. 4
D. 2

Answer

To solve the equation \( 16qxy = (2x + y)^2 - (2x - y)^2 \), we will start by simplifying the right-hand side.

### Step 1: Expand the squares on the right-hand side
We will expand both \((2x + y)^2\) and \((2x - y)^2\).

\[ (2x + y)^2 = (2x)^2 + 2(2x)(y) + y^2 = 4x^2 + 4xy + y^2 \]

\[ (2x - y)^2 = (2x)^2 - 2(2x)(y) + y^2 = 4x^2 - 4xy + y^2 \]

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

\[ 16qxy = (4x^2 + 4xy + y^2) - (4x^2 - 4xy + y^2) \]

### Step 3: Simplify the right-hand side
Now, we simplify the right-hand side:

\[ 16qxy = (4x^2 + 4xy + y^2) - (4x^2 - 4xy + y^2) \]

The \(4x^2\) and \(y^2\) terms cancel out:

\[ 16qxy = 4xy + 4xy = 8xy \]

### Step 4: Equate both sides
Now we have:

\[ 16qxy = 8xy \]

### Step 5: Divide both sides by \(xy\) (assuming \(xy \neq 0\))
We can divide both sides by \(xy\):

\[ 16q = 8 \]

### Step 6: Solve for \(q\)
Now, we solve for \(q\):

\[ q = \frac{8}{16} = \frac{1}{2} \]

Thus, the value of \(q\) is:

\[ \boxed{\frac{1}{2}} \]