If one root is common is equations `x^(2)-ax+b=0 and x^(2 ) +bx-a=0 `, then :

Options

A. `a=b`
B. `a-b=1`
C. `a+b=1`
D. None of these

Answer

To solve the problem, we need to find the relationship between \( a \) and \( b \) given that there is a common root in the two quadratic equations:

1. \( x^2 - ax + b = 0 \)
2. \( x^2 + bx - a = 0 \)

### Step-by-Step Solution:

**Step 1: Let the common root be \( y \).**

Since \( y \) is a root of both equations, we can substitute \( y \) into both equations.

**Step 2: Substitute \( y \) into the first equation.**

From the first equation:
\[ y^2 - ay + b = 0 \tag{1} \]

**Step 3: Substitute \( y \) into the second equation.**

From the second equation:
\[ y^2 + by - a = 0 \tag{2} \]

**Step 4: Subtract equation (1) from equation (2).**

Subtracting equation (1) from equation (2):
\[ (y^2 + by - a) - (y^2 - ay + b) = 0 \]
This simplifies to:
\[ by - a + ay - b = 0 \]
Combining like terms gives:
\[ (b + a)y - (a + b) = 0 \]

**Step 5: Factor out the common terms.**

We can factor this equation:
\[ (b + a)y = a + b \]

**Step 6: Solve for \( y \).**

If \( b + a \neq 0 \), we can divide both sides by \( b + a \):
\[ y = 1 \]

**Step 7: Substitute \( y = 1 \) back into one of the original equations.**

Substituting \( y = 1 \) into the first equation:
\[ 1^2 - a(1) + b = 0 \]
This simplifies to:
\[ 1 - a + b = 0 \]
Rearranging gives:
\[ a - b = 1 \tag{3} \]

**Step 8: Identify the correct option.**

From equation (3), we have:
\[ a - b = 1 \]
This means that the correct option is:
**Option B: \( a - b = 1 \)**.