Question

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

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

Answer 1

The correct option is A $q^2-p^2$

Given:

$\alpha ,\beta$ are the roots of the equation $x^2+px+1=0$ and $\gamma ,\delta$ the roots of the equation, $x^2+qx+1=0$

Now,

$\text{sum of roots}=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

$\alpha +\beta =-p\dots \left(1\right)$

$\gamma +\delta =-q\dots \left(2\right)$

$\text{Product of roots}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$

$\alpha \beta =1\dots \left(3\right)$

$\gamma \delta =1\dots \left(4\right)$

Now,

$(\alpha -\gamma )(\alpha +\delta )(\beta -\gamma )(\beta +\delta )$

$=(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$

$=(\gamma -\alpha )(\gamma -\beta )(\delta +\alpha )(\delta +\beta )$

$=[{\gamma }^2-(\alpha +\beta )\gamma +\alpha \beta ][{\delta }^2+(\alpha +\beta )\delta +\alpha \beta ]$

$=[{\gamma }^2+p\gamma +1][{\delta }^2-p\delta +1]\dots \left(5\right)$

Since $\gamma ,\delta$ are the roots of the equation

$x^2+qx+1=0$

${\gamma }^2+q\gamma +1=0$ and ${\delta }^2+q\delta +1=0$

$\Rightarrow {\gamma }^2+1=-q\gamma$
and ${\delta }^2+1=-q\delta \dots \left(6\right)$

Using this in equation $\left(5\right)$, we get

$(\alpha -\gamma )(\alpha +\delta )(\beta -\gamma )(\beta +\delta )$

$=\left[\right({\gamma }^2+1)+p\gamma ]\left[\right({\delta }^2+1)-p\delta ]$

$=[-q\gamma +p\gamma ][-q\delta -p\delta ]$ (from (6))

$=-\gamma \delta (p-q)(p+q)$

$=-(p^2-q^2)$ (from (4))

$\therefore (\alpha -\gamma )(\alpha +\delta )(\beta -\gamma )(\beta +\delta )=q^2-p^2$

Hence, option (A) is correct.