Question

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

• A. $a-b-c-d$
• B. $a+b+c-d$
• C. $a+b+c+d$
• D. $a+b-c-d$

Answer 1

Answer: (C), (D)

$a+b-c-d$
$a+b+c+d$

We have,

$a+b=-p,ab=1,c+d=-q$ and $cd=1$.

Now, $(a-c)(b-c)(a+d)(b+d)$

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

$=(ab+ad-bc-cd)(ba+db-ca-cd)$

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

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

$=(c+d{)}^2-(a+b{)}^2[\because ab=cd=1]$

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

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

Hence, $(a-c)(b-c)(a+d)(b+d)$ is divisible by $a+b+c+d$ and $a+b-c-d$.