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

Since $\gamma$ and $\delta$ are the roots of the quadratic equation $x^2 + px + r = 0$, any number $x$ satisfying this polynomial equation must be a root. We evaluate the expression $(x - \gamma)(x - \delta)$ by expanding it to $x^2 - (\gamma + \delta)x + \gamma\delta$. Using Vieta's formulas for $x^2 + px + r = 0$, we know the sum of the roots $\gamma + \delta = -p$ and the product of the roots $\gamma\delta = r$. Substituting these values gives the identity $x^2 + px + r$. Since $\alpha$ is given as a root of $x^2 + px - q = 0$, we know that $\alpha^2 + p\alpha - q = 0$, which rearranges to $\alpha^2 + p\alpha = q$. Substituting $x = \alpha$ into our identity yields $(\alpha - \gamma)(\alpha - \delta) = \alpha^2 + p\alpha + r = q + r$. Therefore, the value of the expression is $q + r$.

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