Question

Let $r,s,t,u$ be the roots of the equation $x^4+A{x}^3+B{x}^2+Cx+D=0$. If $rs=tu$, then

• A. $A^{2}D=C^2$
• B. $AD=C^2$
• C. $AD=C$
• D. none of these

Answer 1

Answer: (A)

Let $rs=tu=K$
Let $r+s=P$ ; $t+u=Q$
$\therefore x^4+A{x}^3+B{x}^2+Cx+D$
= $(x^2-Px+K)(x^2-Qx+K)$
$\therefore A=-(P+Q)$ ; $B=PQ+2K$
$C=-(P+Q)K$ ; $D=K^2$
$\therefore \frac{C}{A}=K$
$\Rightarrow \frac{C^2}{A^2}=K^2=D$
$\therefore A^{2}D=C^2$.