Question

If $\alpha ,\beta$ and $\gamma$ are the roots of equation $x^3-3x^2+x+5=0,$ then $y=\Sigma {\alpha }^2+\alpha \beta \gamma$ satisfies the equation.

• A. $y^3+y+2=0$
• B. $y^3-y^2-y-2=0$
• C. $y^3+3y^2-y-3=0$
• D. $y^2+4y^2+5y+20=0$

Answer 1

Answer: (B)

Given $\alpha ,\beta ,\gamma$ are the roots of equation
$x^3-3x^2+x+5=0$
$\therefore$ $\alpha +\beta +\gamma =3$
$\alpha \beta +\beta \gamma +\gamma \alpha =1$
and $\alpha \beta \gamma =-5$
Now, $y=\Sigma {\alpha }^2+\alpha \beta \gamma$
$={\alpha }^2+{\beta }^2+{\gamma }^2+\alpha \beta \gamma$
$={(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )+\alpha \beta \gamma$
$={(3)}^2-2(1)-5$
$\Rightarrow$ $y=2$
So, $y=2$
satisfies the equation $y^3-y^2-y-2=0$