If quadratic equations $x^2+ax+b=0$ and $x^2+px+q=0$ share one similar root then find quadratic equation for which has roots of other roots of both quadratic equations .
1. Let the common root of the quadratic equations \(x^2 + ax + b = 0\) and \(x^2 + px + q = 0\) be \(\alpha\).
2. Since \(\alpha\) is a root of both equations, we have:
   \[
   \alpha^2 + a\alpha + b = 0 \quad \text{and} \quad \alpha^2 + p\alpha + q = 0
   \]
3. Subtracting the second equation from the first, we get:
   \[
   (\alpha^2 + a\alpha + b) - (\alpha^2 + p\alpha + q) = 0
   \]
   Simplifying, we obtain:
   \[
   a\alpha + b - p\alpha - q = 0 \implies (a - p)\alpha = q - b \implies \alpha = \frac{b - q}{p - a}
   \]
4. Let \(\beta\) be the other root of the first equation \(x^2 + ax + b = 0\), and \(\gamma\) be the other root of the second equation \(x^2 + px + q = 0\).
5. Using Vieta's formulas, we know:
   \[
   \alpha + \beta = -a \quad \text{and} \quad \alpha\beta = b
   \]
   \[
   \alpha + \gamma = -p \quad \text{and} \quad \alpha\gamma = q
   \]
6. We need to find the quadratic equation whose roots are \(\beta\) and \(\gamma\).
7. The sum of the roots \(\beta\) and \(\gamma\) is:
   \[
   \beta + \gamma = (\alpha + \gamma) - \alpha = -p - \alpha
   \]
   Substituting \(\alpha = \frac{b - q}{p - a}\), we get:
   \[
   \beta + \gamma = -p - \frac{b - q}{p - a}
   \]
8. The product of the roots \(\beta\) and \(\gamma\) is:
   \[
   \beta\gamma = \frac{b}{\alpha} \cdot \frac{q}{\alpha} = \frac{bq}{\alpha^2}
   \]
   Substituting \(\alpha = \frac{b - q}{p - a}\), we get:
   \[
   \beta\gamma = \frac{bq}{\left(\frac{b - q}{p - a}\right)^2} = \frac{bq(p - a)^2}{(b - q)^2}
   \]
9. Therefore, the required quadratic equation is:
   \[
   x^2 - (\beta + \gamma)x + \beta\gamma = x^2 - \left(-p - \frac{b - q}{p - a}\right)x + \frac{bq(p - a)^2}{(b - q)^2}
   \]

The final answer is \( \boxed{ x^2 - \left(-p - \frac{b - q}{p - a}\right)x + \frac{bq(p - a)^2}{(b - q)^2} } \)