Question

If $\alpha$ and $\beta$ are roots of $x^2+px+1=0$, and $\gamma$ and $\delta$ are the roots of $x^2+qx+1=0$ show that $q^2-p^2-(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )=0$

Answer 1

Given,

$x^2+qx+1=0\to$ roots $\gamma ,\delta$

$x^2+px+1=0\to$ roots $\alpha ,\beta$

Now,

$\gamma +\delta =-q$ and $\gamma \delta =1$..........(1)

$\alpha +\beta =-p$ and $\alpha \beta =1$..........(2)

Here,

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

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

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

$=[1+p^{\gamma }+{\gamma }^2][1-p^{\delta }+{\delta }^2]$

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

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

$=-1(p^2-q^2)=q^2-p^2$

Now, $q^2-p^2-(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$

$\Rightarrow q^2-p^2-[q^2-p^2]$

$\Rightarrow q^2-p^2-q^2+p^2$

$\Rightarrow 0$

Hence, zero is the correct answer.