Question

Let $\alpha ,\beta ,\gamma$be the roots of the equations,$x^3+a{x}^2+bx+c=0,$ ($a,b,c\in R$and $a,b\&a,b\ne 0$). The system of the equations (in $u,v,w$) given by $\alpha u+\beta v+\gamma w=0;\beta u+\gamma v+\alpha w=0;\gamma u+\alpha v+\beta w=0$ has non-trivial solutions, then the value of $\frac{a^2}{b}$ is

• A. $5$
• B. $1$
• C. $0$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

Finding the value of $\frac{a^2}{b}$:

Given that $\alpha ,\beta ,\gamma$be the roots of the equations,$x^3+a{x}^2+bx+c=0,$

Therefore, $\alpha +\beta +\gamma =-a;\alpha \beta +\beta \gamma +\gamma \alpha =b$

We know that for non-trivial solution

$\begin{array}{l}\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0\end{array}$

$$\begin{gathered} \Rightarrow {\alpha }^3+{\beta }^3+{\gamma }^3–3\alpha \beta \gamma =0 \\ \Rightarrow (\alpha +\beta +\gamma )[{(\alpha +\beta +\gamma )}^2-3(\alpha \beta +\beta \gamma +\gamma \alpha \left)\right]=0 \\ \Rightarrow -a(a^2-3b)=0\left[\because \alpha +\beta +\gamma =-a;\alpha \beta +\beta \gamma +\gamma \alpha =b\right] \\ \Rightarrow (a^2-3b)=0 \\ \Rightarrow a^2=3b \\ \Rightarrow \frac{a^2}{b}=3 \end{gathered}$$

Hence, option (D) is correct.