If  $\alpha ;\beta$    be the roots of $x^2+px+q=0$ and  $y,\delta$  be the roots of $x^2+px+r=0$ then $\frac{(\alpha -\gamma )(\alpha -\delta )}{(\beta -\gamma )(\beta -\delta )}$ 

 is equal to 

Given, a and /3 are the roots of $x^2+px-q=0 .......( i )$

$\begin{array}{l}\alpha +\beta =-p ......( ii ) \\ \alpha \beta =-q .... ( iii) \end{array}$

and  $\text{ Y. }\delta$   are the roots of the equation

 $\begin{array}{r}\therefore x^2+px+r=0\\ \Rightarrow \gamma +\delta =-p\\ \gamma \delta =r\end{array}$

 Now,  $(\alpha -\gamma )(\alpha -\delta )={\alpha }^2-(\gamma +8)\alpha +\gamma \delta$

 $\begin{array}{l}\Rightarrow (\alpha -\gamma )(\alpha -\delta )={\alpha }^2+p\alpha +r\\ \Rightarrow (\alpha -\gamma )(\alpha -\delta )=q+r\\ \left[\because \alpha \text{ is root of }{x}^2+px-q=0\Rightarrow {\alpha }^2+p\alpha -q=0\right.\\ \left.\Rightarrow {\alpha }^2+p\alpha =q\right]\end{array}$

and    $(\beta -\gamma )(\beta -\delta )={\beta }^2-(\gamma +\delta )\beta +\gamma \delta ={\beta }^2+\beta \beta +r$

    $\begin{array}{r}\left[\because \beta \text{ is root of }{x}^2+\rho x-q=0\Rightarrow {\beta }^2+\right.\rho \beta -q=0\\ \left.\to {\beta }^2+p\beta =q\right]\end{array}$

  $\Rightarrow (\beta -\gamma )(\beta -\delta )=q+r$

 On dividing Eq. (vi) by Eq. (vii), we get  

 $\frac{(\alpha -\gamma )(\alpha -\delta )}{(\beta -\gamma )(\beta -\delta )}=\frac{q+r}{q+r}=1$