Step 1: Understanding the Concept:
This is an algebraic equation involving squared binomials. We need to solve for the variable \(x\).
Step 2: Key Formula or Approach:
There are two main approaches:
1. Expand both sides of the equation using the formula \( (a-b)^2 = a^2 - 2ab + b^2 \) and then solve the resulting equation.
2. Take the square root of both sides, remembering to account for both positive and negative roots.
Step 3: Detailed Explanation:
Method 1: Expanding the squares
We are given the equation \( (x-1)^2 = (x-2)^2 \).
Expand the left side: \( (x-1)^2 = x^2 - 2(x)(1) + 1^2 = x^2 - 2x + 1 \).
Expand the right side: \( (x-2)^2 = x^2 - 2(x)(2) + 2^2 = x^2 - 4x + 4 \).
Now, set the expanded forms equal to each other:
\[ x^2 - 2x + 1 = x^2 - 4x + 4 \]
The \(x^2\) terms on both sides cancel each other out. We can subtract \(x^2\) from both sides:
\[ -2x + 1 = -4x + 4 \]
Now, we solve this linear equation. Add \(4x\) to both sides:
\[ -2x + 4x + 1 = 4 \]
\[ 2x + 1 = 4 \]
Subtract 1 from both sides:
\[ 2x = 3 \]
Divide by 2:
\[ x = \frac{3}{2} \]
Method 2: Taking the square root
If \(a^2 = b^2\), then \(a = b\) or \(a = -b\).
Case 1: \( x-1 = x-2 \)
Subtracting \(x\) from both sides gives \( -1 = -2 \), which is false. So there is no solution in this case.
Case 2: \( x-1 = -(x-2) \)
\[ x-1 = -x + 2 \]
Add \(x\) to both sides:
\[ 2x - 1 = 2 \]
Add 1 to both sides:
\[ 2x = 3 \]
Divide by 2:
\[ x = \frac{3}{2} \]
Both methods yield the same result.
Step 4: Final Answer:
The solution to the equation is \( x = \frac{3}{2} \). This corresponds to option (D). (Note: The provided image shows options A, B, C, but D and E are standard for a 5-option question and are assumed for completeness).