If the roots of `x^(2)-bx+c=0` be two consecutive integers, then find the value of `b^(2)-4c`.

Answer

To solve the problem, we need to find the value of \( b^2 - 4c \) given that the roots of the quadratic equation \( x^2 - bx + c = 0 \) are two consecutive integers.

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the equation be \( \alpha \) and \( \beta \). Since they are consecutive integers, we can express them as \( \alpha = n \) and \( \beta = n + 1 \) for some integer \( n \).

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for the quadratic equation \( x^2 - bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = b \)
- The product of the roots \( \alpha \cdot \beta = c \)

Therefore, we can write:
\[ n + (n + 1) = b \implies 2n + 1 = b \]
\[ n(n + 1) = c \implies c = n^2 + n \]

3. **Substitute \( b \) and \( c \)**:
Now we have expressions for \( b \) and \( c \):
- \( b = 2n + 1 \)
- \( c = n^2 + n \)

4. **Calculate \( b^2 - 4c \)**:
We need to find \( b^2 - 4c \):
\[ b^2 = (2n + 1)^2 = 4n^2 + 4n + 1 \]
\[ 4c = 4(n^2 + n) = 4n^2 + 4n \]
Now substituting these into the expression:
\[ b^2 - 4c = (4n^2 + 4n + 1) - (4n^2 + 4n) \]
Simplifying this gives:
\[ b^2 - 4c = 1 \]

5. **Final Result**:
Thus, the value of \( b^2 - 4c \) is \( 1 \).