Find the number of real roots of the equation `(x-1)^2+(x-2)^2+(x-3)^2=0.`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of real roots of the equation \((x-1)^2 + (x-2)^2 + (x-3)^2 = 0\), we will follow these steps:

### Step 1: Understand the equation
The equation consists of three squared terms. Since the square of any real number is always non-negative, each term \((x-1)^2\), \((x-2)^2\), and \((x-3)^2\) is greater than or equal to zero.

### Step 2: Analyze when the sum equals zero
For the sum of these squares to equal zero, each individual square must also be zero. This means:
\[ (x-1)^2 = 0, \quad (x-2)^2 = 0, \quad (x-3)^2 = 0 \]

### Step 3: Solve each equation
1. From \((x-1)^2 = 0\):
\[ x - 1 = 0 \implies x = 1 \]
2. From \((x-2)^2 = 0\):
\[ x - 2 = 0 \implies x = 2 \]
3. From \((x-3)^2 = 0\):
\[ x - 3 = 0 \implies x = 3 \]

### Step 4: Check if all conditions are satisfied
For the original equation to hold true, all three conditions must be satisfied simultaneously. However, there is no single value of \(x\) that can satisfy all three equations at the same time. Therefore, we cannot have a real number \(x\) that makes the entire left-hand side equal to zero.

### Conclusion
Since the only way for the sum of squares to equal zero is for each square to be zero, and since no single \(x\) can satisfy all three conditions, we conclude that there are no real roots for the equation.

### Final Answer
The number of real roots of the equation \((x-1)^2 + (x-2)^2 + (x-3)^2 = 0\) is **0**.

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