Question

If the difference between the roots of $ {{x}^{2}}+ax-b=0 $ is equal to the difference between the roots of $ {{x}^{2}}-px+q=0, $ then $ {{a}^{2}}-{{p}^{2}} $ in terms of $b$ and $q$ is

• A. $ -4(b+q) $
• B. $ 4(b+q) $
• C. $ 4(b-q) $
• D. $ 4(q-b) $

Answer 1

Answer: (A)

Let $ \alpha ,\beta $ are the roots of the equation
$ {{x}^{2}}+ax-b=0 $
$ \therefore $ $ \alpha +\beta =-a,\,\,\alpha \beta =-b $
and $ \gamma ,\delta $
are the roots of the equation
$ {{x}^{2}}-px+q=0 $
$ \therefore $ $ \gamma +\delta =p,\,\gamma \delta =q $
Given, $ \alpha -\beta =\gamma -\delta $
$ \Rightarrow $ $ {{(\alpha -\beta )}^{2}}={{(\gamma -\delta )}^{2}} $
$ \Rightarrow $ $ {{(\alpha +\beta )}^{2}}-4\alpha \beta ={{(\gamma +\delta )}^{2}}-4\gamma \delta $
$ \Rightarrow $ $ {{a}^{2}}+4b={{p}^{2}}-4q $
$ \Rightarrow $ $ {{a}^{2}}-{{p}^{2}}=-4(b+q) $