Question

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

• A. $p^2-q^2$
• B. $p^2-2q$
• C. $p^2+q^2$
• D. $2(p^2-q^2)$

Answer 1

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

Since $a$ and $b$ are the roots of the equation $x^2+px+1=0$

$\therefore a+b=-p$ ...(1)

and $ab=1$ ...(2)

Also, since $c$ and $d$ are the roots of the equation $x^2+qx+1=0$

$\therefore c+d=-q$ ...(3)

and, $cd=1$ ...(4)

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

$=(ab-bc-ac+c^2)(ab+db+ad+d^2)$

$=[ab-c(b+a)+c^2].[ab+d(a+b)+d^2]=(1+cp+c^2)(1-pd+d^2)$

$[$Putting the values of $a+b$ and $ab]$

$=1+cp+c^2-pd-cd{p}^2-c^{2}pd+d^2+cp{d}^2+c^2{d}^2$

$=1+(c^2+d^2)+c^2{d}^2-cd{p}^2+p(c-d)+cpd(d-c)$

$=1+[{(c+d)}^2-2cd]+c^2{d}^2-cd{p}^2+p(c-d)+cpd(d-c)$

$=1+(q^2-2)+1-p^2+p(c-d)+p(d-c)$

$[$Putting the values of $c+d$ and $cd]$

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