Solve
`4x^(2)-4ax+(a^(2)-b^(2))=0.`

Options

A. `((-a+b))/(2)` and `((a-b))/(2)`
B. `((a+b))/(2)` and `((a-b))/(2)`
C. `((a+b))/(2)` and `((-a-b))/(2)`
D. none of these

Explanation

To solve the quadratic equation \(4x^2 - 4ax + (a^2 - b^2) = 0\) using the factorization method, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(Ax^2 + Bx + C = 0\), where:
- \(A = 4\)
- \(B = -4a\)
- \(C = a^2 - b^2\)

### Step 2: Factor the constant term
We know that \(a^2 - b^2\) can be factored using the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]

### Step 3: Rewrite the equation
Now, we can rewrite the equation as:
\[ 4x^2 - 4ax + (a - b)(a + b) = 0 \]

### Step 4: Factor by grouping
We need to express the middle term \(-4ax\) in a way that allows us to factor the equation. We can rewrite \(-4ax\) as:
\[ -2a \cdot 2x + 2b \cdot 2x = -2a \cdot 2x + 2b \cdot 2x \]
This gives us the equation:
\[ 4x^2 - 2a \cdot 2x + 2b \cdot 2x + (a - b)(a + b) = 0 \]

### Step 5: Group the terms
Now we can group the terms:
\[ (4x^2 - 2a \cdot 2x) + (2b \cdot 2x + (a - b)(a + b)) = 0 \]

### Step 6: Factor out common terms
From the first group, we can factor out \(2x\):
\[ 2x(2x - 2a) + (a - b)(a + b) = 0 \]

### Step 7: Rewrite in factored form
Now we can factor the entire expression:
\[ 2x(2x - (a - b))(2x - (a + b)) = 0 \]

### Step 8: Set each factor to zero
Now we set each factor to zero:
1. \(2x - (a - b) = 0\)
2. \(2x - (a + b) = 0\)

### Step 9: Solve for \(x\)
From the first equation:
\[ 2x = a - b \implies x = \frac{a - b}{2} \]
From the second equation:
\[ 2x = a + b \implies x = \frac{a + b}{2} \]

### Final Solution
Thus, the solutions to the equation \(4x^2 - 4ax + (a^2 - b^2) = 0\) are:
\[ x = \frac{a - b}{2} \quad \text{and} \quad x = \frac{a + b}{2} \]

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