Solve this following


Let $\alpha, \beta, \gamma$ be the real roots of the equation, $x^{3}+a x^{2}+b x+c=0,(a, b, c \in R$ and $a, b \neq 0)$. If the system of equations (in, $\mathrm{u}, \mathrm{v}, \mathrm{w}$ ) given by $\alpha u+\beta v+\gamma w=0, \beta u+\gamma v+\alpha w=0 ;$ $\gamma \mathrm{u}+\alpha \mathrm{v}+\beta \mathrm{w}=0$ has non-trivial solution, then the

value of $\frac{\mathrm{a}^{2}}{\mathrm{~b}}$ is


  1. 5

  2. 3

  3. 1

  4. 0

Correct Option: , 2


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

$\Rightarrow-(\alpha+\beta+\gamma)\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\sum \alpha \beta\right)=0$

$\Rightarrow-(-a)\left(a^{2}-2 b-b\right)=0$

$\Rightarrow a\left(a^{2}-3 b\right)=0$

$\Rightarrow a^{2}=3 b \Rightarrow \frac{a^{2}}{b}=3$


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