Question

If $\alpha ,\beta$ are the roots of the equation $x^2+px+1=0;\gamma ,\delta$ the roots of the equation $x^2+px+1=0;$ then $(\alpha -\gamma )(\alpha +\delta )(\beta -\gamma )(\beta +\delta )$

• A. $q^2-p^2$
• B. $p^2-q^2$
• C. $p^2+q^2$
• D. none of these

Answer 1

The correct option is A

$q^2-p^2$

$Given:\alpha ,\beta aretherootsoftheequation{x}^2+px+1=0Also,\gamma and\delta aretherootsoftheequation{x}^2+qx+1=0\text{Then, the sum and the product of the roots of the given equation are as follows}:\alpha +\beta =-\frac{p}{1}=-p\alpha \beta =\frac{1}{1}=1\gamma +\delta =-\frac{q}{1}=-q\gamma \delta =\frac{1}{1}=1Moreover,(\gamma +\delta {)}^2={\gamma }^2+{\delta }^2+2\gamma \delta \Rightarrow {\gamma }^2+{\delta }^2=q^2-2\therefore (\alpha -\gamma \left)\right(\alpha +\delta \left)\right(\beta -\gamma \left)\right(\beta +\delta )=(\alpha -\gamma \left)\right(\beta -\delta \left)\right(\alpha +\delta \left)\right(\beta +\delta )=(\alpha \beta -\alpha \gamma -\beta \gamma +{\gamma }^2\left)\right(\alpha \beta +\alpha \delta +\beta \delta +{\delta }^2)=[\alpha \beta -\gamma (\alpha +\beta )+{\gamma }^2\left]\right(\alpha \beta +\delta (\alpha +\beta )+{\delta }^2]=(1-\gamma (-p)+{\gamma }^2\left)\right(1+\delta (-p)+{\delta }^2)=(1+\gamma p+{\gamma }^2\left)\right(1-\delta p+{\delta }^2)=1-p\delta +{\delta }^2+p\gamma -p^2\gamma \delta +p\gamma {\delta }^2+{\gamma }^2-p\delta {\gamma }^2+{\gamma }^2{\delta }^2=1-p\delta +p\gamma +{\delta }^2-p^2\gamma \delta +p\gamma {\delta }^2+{\gamma }^2-p\delta {\gamma }^2+{\gamma }^2{\delta }^2=1-p(\delta -\gamma )-p^2\gamma \delta +p\gamma \delta (\delta -\gamma )+({\gamma }^2+{\delta }^2+1)=1-p^2\gamma \delta +p\gamma \delta (\delta -\gamma )-p(\delta -\gamma )+({\gamma }^2+{\delta }^2)+1=1-p^2(\delta -\gamma \left)p\right(\gamma \delta -1)+q^2-2+1=-p^2+(\delta -\gamma \left)p\right(1-l)+q^2=q^2-p^2$