If the roots of `x^2-bx +c=0` are two consecutive integers then `b^2-4c=`

Explanation

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. **Define the Roots:**
Let the two consecutive integers be \( \alpha \) and \( \alpha + 1 \).

2. **Sum of the Roots:**
According to Vieta's formulas, the sum of the roots of the quadratic equation \( x^2 - bx + c = 0 \) is given by:
\[ \alpha + (\alpha + 1) = b \]
Simplifying this, we get:
\[ 2\alpha + 1 = b \quad \text{(Equation 1)} \]

3. **Product of the Roots:**
The product of the roots is also given by Vieta's formulas:
\[ \alpha(\alpha + 1) = c \]
Expanding this, we have:
\[ \alpha^2 + \alpha = c \quad \text{(Equation 2)} \]

4. **Express \( \alpha \) in terms of \( b \):**
From Equation 1, we can express \( \alpha \):
\[ 2\alpha = b - 1 \implies \alpha = \frac{b - 1}{2} \]

5. **Substitute \( \alpha \) into Equation 2:**
Now substitute \( \alpha \) into Equation 2:
\[ \left(\frac{b - 1}{2}\right)^2 + \left(\frac{b - 1}{2}\right) = c \]
Simplifying this:
\[ \frac{(b - 1)^2}{4} + \frac{b - 1}{2} = c \]
To combine the terms, convert \( \frac{b - 1}{2} \) into a fraction with a denominator of 4:
\[ \frac{(b - 1)^2}{4} + \frac{2(b - 1)}{4} = c \]
Thus:
\[ \frac{(b - 1)^2 + 2(b - 1)}{4} = c \]
Simplifying the numerator:
\[ (b - 1)^2 + 2(b - 1) = b^2 - 2b + 1 + 2b - 2 = b^2 - 1 \]
Therefore:
\[ c = \frac{b^2 - 1}{4} \]

6. **Calculate \( b^2 - 4c \):**
Now, we need to find \( b^2 - 4c \):
\[ b^2 - 4c = b^2 - 4\left(\frac{b^2 - 1}{4}\right) \]
This simplifies to:
\[ b^2 - (b^2 - 1) = b^2 - b^2 + 1 = 1 \]

### Final Answer:
Thus, the value of \( b^2 - 4c \) is:
\[ \boxed{1} \]