Question

The condition that the product of two roots of $a{x}^3+b{x}^2+cx+d=0$ may be equal to $-1$, is:

• A. $c(a+d)+b(a+d)=0$
• B. $d(b+d)+a(a+c)=0$
• C. $a(b+c)+b(c+d)=0$
• D. $a(a+b)+d(a+c)=0$

Answer 1

The correct option is B $d(b+d)+a(a+c)=0$

Let $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$
Such that $\alpha \beta =-1$
$S_1=\alpha +\beta +\gamma =-\frac{-b}{a}\Rightarrow \alpha +\beta =\frac{-b}{a}-\gamma S_2=\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}\Rightarrow \gamma (\alpha +\beta )=\frac{c}{a}+1=\frac{c+a}{a}{S}_3=\alpha \beta \gamma =-\frac{d}{a}\Rightarrow \gamma =\frac{d}{a}$
From $S_1$ and $S_3$,
$\alpha +\beta =\frac{-b}{a}-\frac{d}{a}=\frac{-(b+d)}{a}$ (4)
From $S_2$ and (4), we have
$\frac{-(b+d)}{a}\left(\frac{d}{a}\right)=\frac{c+a}{a}\Rightarrow d(b+d)+a(a+c)=0$
Hence, option 'B' is correct.