If $\alpha ,\beta $ are the roots of the equation ${x^2} + px - q = 0$ and $\gamma ,\delta $ are the roots of the equation ${x^2} + px + r = 0$, then the value of $(\alpha - \gamma )(\alpha - \delta )$ is
$
  A)p + r \\
  B)p - r \\
  C)q - r \\
  D)q + r \\
 $

Hint: Here to solve this problem we get the given value we need to know the sum of the roots and products of roots of given quadratic equations. They have mentioned the roots of given equations.

Complete step-by-step answer:

We know that if there is any quadratic equation in the form $a{x^2} + bx + c = 0$.Then we know that
Sum of roots $ = -\dfrac{{{\text{ coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Product of roots $ = \dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$
Here given equation is ${x^2} + px - q = 0$ whose roots are $\alpha ,\beta $

Now by using the above concept
Sum of the roots $\left( {\alpha + \beta } \right)$ $ = \dfrac{{ - p}}{1} = - p$
Product of roots $\left( {\alpha .\beta } \right) = \dfrac{{ - q}}{1} = - q$
Here it is also mentioned $\gamma ,\delta $ are the root of equation ${x^2} + px + r = 0$.
Sum of the roots $\left( {\gamma + \delta } \right) = \dfrac{{ - p}}{1} = - p$
Product of roots $\left( {\gamma \delta } \right) = \dfrac{r}{1} = r$
Now here we have to find the value of $(\alpha - \gamma )(\alpha - \delta )$
$ \Rightarrow \left( {\alpha - \gamma } \right)\left( {\alpha - \delta } \right)$
$ \Rightarrow {\alpha ^2} - \alpha (\gamma + \delta ) + \gamma \delta $
Since we the value $\gamma + \delta = - p$ and $\gamma \delta = r$.So now on substituting the value and further simplification we get
$ \Rightarrow q + r$
Option D is the correct one .

Note: In this problem before finding the value of given term we have got the required value of to solve the given term. So get the value we have used the sum of roots and products of roots of the given two functions. Here we have to concentrate on finding the sum of roots of the equation and product roots of the equation by using the formulas.