The equations `x^2+b^2=1-2bx` and `x^2+a^2=1-2ax` have only oneroot in common then `|a-b|=`

Options

A. 1
B. 0
C. 2
D. none of these

Explanation

To solve the problem, we start with the two given equations:

1. \( x^2 + b^2 = 1 - 2bx \)
2. \( x^2 + a^2 = 1 - 2ax \)

We need to manipulate these equations to find the absolute value of \( |a - b| \) given that they have only one root in common.

### Step 1: Rearranging the equations

We can rearrange both equations to standard quadratic form:

For the first equation:
\[ x^2 + 2bx + b^2 - 1 = 0 \]
For the second equation:
\[ x^2 + 2ax + a^2 - 1 = 0 \]

### Step 2: Identifying the roots

The roots of these quadratic equations can be found using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

For the first equation, the roots are:
\[ x = \frac{-2b \pm \sqrt{(2b)^2 - 4(1)(b^2 - 1)}}{2(1)} = \frac{-2b \pm \sqrt{4b^2 - 4(b^2 - 1)}}{2} = \frac{-2b \pm \sqrt{4}}{2} = -b \pm 1 \]
Thus, the roots are:
\[ x = 1 - b \quad \text{and} \quad x = -1 - b \]

For the second equation, the roots are:
\[ x = \frac{-2a \pm \sqrt{(2a)^2 - 4(1)(a^2 - 1)}}{2(1)} = \frac{-2a \pm \sqrt{4a^2 - 4(a^2 - 1)}}{2} = \frac{-2a \pm \sqrt{4}}{2} = -a \pm 1 \]
Thus, the roots are:
\[ x = 1 - a \quad \text{and} \quad x = -1 - a \]

### Step 3: Setting up the conditions for a common root

Since the two equations have only one root in common, we can set up the following conditions:

1. \( 1 - b = 1 - a \)
2. \( 1 - b = -1 - a \)
3. \( -1 - b = 1 - a \)

### Step 4: Solving the conditions

**Condition 1:**
\[ 1 - b = 1 - a \implies a = b \]
This leads to both equations being the same, which contradicts the problem statement that they have only one root in common. Therefore, this condition is not valid.

**Condition 2:**
\[ 1 - b = -1 - a \implies 1 + a = 1 - b \implies a + b = -2 \implies |a - b| = 2 \]

**Condition 3:**
\[ -1 - b = 1 - a \implies a - b = 2 \implies |a - b| = 2 \]

### Conclusion

From both valid conditions, we find that:
\[ |a - b| = 2 \]