If the difference of the roots of `x^2-p x+q=0` is unity, then
a) `p^2+4q=1` b) `p^2-4q=1` c) `p^2+4q^2=(1+2q)^2` d) `4p^2+q^2=(1+2p)^2`

Explanation

To solve the problem, we need to find the relationship between \( p \) and \( q \) given that the difference of the roots of the quadratic equation \( x^2 - px + q = 0 \) is unity (i.e., 1).

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the equation be \( \alpha \) and \( \beta \). We know that:
\[ \alpha - \beta = 1 \quad \text{(Equation 1)} \]

2. **Use the Sum of Roots**:
From Vieta's formulas, we know:
\[ \alpha + \beta = p \quad \text{(Equation 2)} \]

3. **Express Roots in Terms of \( \alpha \)**:
From Equation 1, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = \alpha - 1 \]

4. **Substitute \( \beta \) in Equation 2**:
Substitute \( \beta \) in Equation 2:
\[ \alpha + (\alpha - 1) = p \]
Simplifying this gives:
\[ 2\alpha - 1 = p \]
Therefore:
\[ 2\alpha = p + 1 \quad \Rightarrow \quad \alpha = \frac{p + 1}{2} \]

5. **Find \( \beta \)**:
Substitute \( \alpha \) back to find \( \beta \):
\[ \beta = \alpha - 1 = \frac{p + 1}{2} - 1 = \frac{p - 1}{2} \]

6. **Calculate \( \alpha^2 + \beta^2 \)**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
We can calculate \( \alpha\beta \) using Vieta's formulas:
\[ \alpha\beta = q \]
Thus:
\[ \alpha^2 + \beta^2 = p^2 - 2q \]

7. **Use the Identity**:
We can also express \( \alpha^2 - \beta^2 \) as:
\[ \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) = 1 \cdot p = p \]
Therefore:
\[ \alpha^2 + \beta^2 = \frac{p^2 + 1}{2} \]

8. **Set Up the Equation**:
Now we have two expressions for \( \alpha^2 + \beta^2 \):
\[ p^2 - 2q = \frac{p^2 + 1}{2} \]

9. **Solve for \( q \)**:
Rearranging gives:
\[ 2(p^2 - 2q) = p^2 + 1 \]
Simplifying this:
\[ 2p^2 - 4q = p^2 + 1 \]
Thus:
\[ p^2 - 4q = 1 \]

10. **Conclusion**:
Therefore, we have derived that:
\[ p^2 - 4q = 1 \]
This corresponds to option (b).

### Final Answer:
The correct option is **(b) \( p^2 - 4q = 1 \)**.