Consider quadratic equations `x^(2)-ax+b=0`……….`(i)` and `x^(2)+px+q=0`……….`(ii)`
If for the equations `(i)` and `(ii)` , one root is common and the equation `(ii)` have equal roots, then `b+q` is equal to

Explanation

To solve the problem, we need to analyze the two quadratic equations given and use the information provided about their roots. Let's break down the solution step by step.

### Step 1: Define the equations and their roots
We have two quadratic equations:
1. \( x^2 - ax + b = 0 \) (Equation 1)
2. \( x^2 + px + q = 0 \) (Equation 2)

Let the roots of Equation 1 be \( \alpha \) and \( \beta \), where \( \alpha \) is the common root with Equation 2. Since Equation 2 has equal roots, both roots of Equation 2 are \( \alpha \).

### Step 2: Use Vieta's formulas for Equation 1
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = a \)
- The product of the roots \( \alpha \beta = b \)

### Step 3: Use Vieta's formulas for Equation 2
For Equation 2, since both roots are equal and equal to \( \alpha \):
- The sum of the roots \( \alpha + \alpha = -p \) which simplifies to \( 2\alpha = -p \) or \( \alpha = -\frac{p}{2} \)
- The product of the roots \( \alpha \cdot \alpha = q \) which gives \( \alpha^2 = q \)

### Step 4: Substitute \( \alpha \) into the expressions for \( b \) and \( q \)
From the product of the roots of Equation 1, we have:
\[ b = \alpha \beta \]
From the sum of the roots of Equation 1, we can express \( \beta \) as:
\[ \beta = a - \alpha \]
Substituting \( \alpha = -\frac{p}{2} \) into the expression for \( \beta \):
\[ \beta = a + \frac{p}{2} \]

Now substituting \( \beta \) into the expression for \( b \):
\[ b = \alpha \beta = \left(-\frac{p}{2}\right) \left(a + \frac{p}{2}\right) \]

### Step 5: Calculate \( b + q \)
Now we need to find \( b + q \):
\[ q = \alpha^2 = \left(-\frac{p}{2}\right)^2 = \frac{p^2}{4} \]
Thus,
\[ b + q = \left(-\frac{p}{2}\right) \left(a + \frac{p}{2}\right) + \frac{p^2}{4} \]
Expanding \( b \):
\[ b = -\frac{p}{2}a - \frac{p^2}{4} \]
So,
\[ b + q = -\frac{p}{2}a - \frac{p^2}{4} + \frac{p^2}{4} \]
The \( \frac{p^2}{4} \) terms cancel out:
\[ b + q = -\frac{p}{2}a \]

### Final Result
Thus, the final result for \( b + q \) is:
\[ b + q = -\frac{p}{2} a \]