Question

If $a,b$ are the roots of $x^2+px+1=0$ and $c,d$ are the roots of $x^2+qx+1=0,$ the value of $E=(a-c)(b-c)(a+d)(b+d)$ is

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

Answer 1

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

$a,b$ roots of $x^2+px+1=0\Rightarrow a+b=-p,ab=1$

$c,d$ roots of $x^2+qx+1=0\Rightarrow c+d=-q,cd=1$

Given,

$E=(a-c)(b-c)(a+d)(b+d)$

$=\left[\right(a-c\left)\right(b+d\left)\right]\left[\right(b-c\left)\right(a+d\left)\right]$

$=(ab+ad-cb-cd)(ab+bd-ac-cd)$

$=(1+ad-cb-1)(1+bd-ac-1)$

$=(ad-cb)(bd-ac)$

$=ab{d}^2-a^{2}dc-b^{2}cd+ab{c}^2$

$=d^2-a^2-b^2+c^2$

$=(c^2+d^2)-(a^2+b^2)$

$=\left[\right(c+d{)}^2-2cd]-[(a+b{)}^2-2ab]$

$=\left[\right(-q{)}^2-2]-[(p{)}^2-2]$

$=q^2-2-p^2+2$

$=q^2-p^2$