Question

If $a,b$ are the roots of $x^2+px+1=0$ and $c,d$ are the roots of $x^2+qx+1=0$. Then $\frac{(a-c)(b-c)(a+d)(b+d)}{(q^2-p^2)}=$.

Answer 1

$x^2+px+1=0$

$a+b=-p$

$ab=1$

$x^2+qx+1=0$

$c+d=-q$

$cd=1$

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

$(1+pc+c^2)$

$(a+d)(b+d)=ab+(a+b)d+d^2$

$=(1-pd+d^2)$

$\therefore (1+pc+c^2)(1-pd+d^2)$

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

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

$=1+c^2+d^2+pc-pc+pd-pd+p^2+1$

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

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

$=p^2+q^2$

$\therefore \frac{(a-c)(b-c)(a+d)(b+d)}{q^2-p^2}$

$=\frac{q^2+p^2}{q^2-p^2}$