Question

Let $r,s,t$ and $u$ be the roots of the equation $x^4+A{x}^3+B{x}^2+Cx+D=0;$
$A,B,C,D\in R.$ If $rs=tu$, then $A^{2}D$ is equal to

• A. $B^2$
• B. $C^2$
• C. $0$
• D. $BC$

Answer 1

The correct option is B $C^2$

Since, $r,s,t$ and $u$ are roots of the equation $x^4+A{x}^3+B{x}^2+Cx+D=0$
We know that,
$r+s+t+u=-A\cdots \left(1\right)$
$(r+s)(t+u)+rs+tu=B\cdots \left(2\right)$
$rs(t+u)+tu(r+s)=-C\cdots \left(3\right)$
$rstu=D\cdots \left(4\right)\Rightarrow (rs{)}^2=D\cdots (5)$
$[\because rs=tu]$

Using equation $\left(3\right)$,
$rs(r+s+t+u)=-C$
Using equation $\left(1\right)$,
$\Rightarrow rs=\frac{C}{A}$
Using equation $\left(5\right)$,
$\Rightarrow \frac{C^2}{A^2}=D\Rightarrow A^{2}D=C^2$