Question

If α, β are the roots of the equation $x^2+px+1=0;\gamma ,\delta$ the roots of the equation $x^2+qx+1=0,then(\mathrm{\alpha }-\mathrm{\gamma })(\mathrm{\alpha }+\mathrm{\delta })(\mathrm{\beta }-\mathrm{\gamma })(\mathrm{\beta }+\mathrm{\delta })=$
(a) $q^2-p^2$
(b) $p^2-q^2$
(c) $p^2+q^2$
(d) none of these.

Answer 1

(a) $q^2-p^2$
Given: $\alpha and\beta$ are the roots of the equation $x^2+px+1=0$.
Also, $\gamma and\delta$ are the roots of the equation $x^2+qx+1=0$.
Then, the sum and the product of the roots of the given equation are as follows:
$$\begin{gathered} \alpha +\beta =-\frac{p}{1}=-p \\ \alpha \beta =\frac{1}{1}=1 \\ \gamma +\delta =-\frac{q}{1}=-q \\ \gamma \delta =\frac{1}{1}=1 \end{gathered}$$


$$\begin{gathered} \mathrm{Moreover,}(\gamma +\delta {)}^2={\gamma }^2+{\delta }^2+2\gamma \delta \\ \Rightarrow {\gamma }^2+{\delta }^2=q^2-2 \end{gathered}$$


$$\begin{gathered} \therefore (\alpha -\gamma )(\alpha +\delta )(\beta -\gamma )(\beta +\delta )=(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta ) \\ =\left(\alpha \beta -\alpha \gamma -\beta \gamma +{\gamma }^2\right)\left(\alpha \beta +\alpha \delta +\beta \delta +{\delta }^2\right) \\ =\left[\alpha \beta -\gamma \left(\alpha +\beta \right)+{\gamma }^2\right]\left[\alpha \beta +\delta \left(\alpha +\beta \right)+{\delta }^2\right] \\ =(1-\gamma (-p)+{\gamma }^2)(1+\delta (-p)+{\delta }^2) \\ =(1+\gamma p+{\gamma }^2)(1-\delta p+{\delta }^2) \\ =1-p\delta +{\delta }^2+p\gamma -p^2\gamma \delta +p\gamma {\delta }^2+{\gamma }^2-p\delta {\gamma }^2+{\gamma }^2{\delta }^2 \\ =1-p\delta +p\gamma +{\delta }^2-p^2\gamma \delta +p\gamma {\delta }^2+{\gamma }^2-p\delta {\gamma }^2+{\gamma }^2{\delta }^2 \\ =1-p(\delta -\gamma )-p^2\gamma \delta +p\gamma \delta (\delta -\gamma )+({\gamma }^2+{\delta }^2)+1 \\ =1-p^2\gamma \delta +p\gamma \delta (\delta -\gamma )-p(\delta -\gamma )+({\gamma }^2+{\delta }^2)+1 \\ =1-p^2+(\delta -\gamma )p(\gamma \delta -1)+q^2-2+1 \\ =-p^2+(\delta -\gamma )p(1-1)+q^2 \\ =q^2-p^2 \end{gathered}$$