JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 3)
Let $$\alpha$$, $$\beta$$, $$\gamma$$ be the real roots of the equation, x3 + ax2 + bx + c = 0, (a, b, c $$\in$$ R and a, b $$\ne$$ 0). If the system of 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 solution, then the value of $${{{a^2}} \over b}$$ is
5
3
1
0
Explanation
x3 + ax2 + bx + c = 0
Roots are $$\alpha$$, $$\beta$$, $$\gamma$$.
For non-trivial solutions,
$$\left| {\matrix{ \alpha & \beta & \gamma \cr \beta & \gamma & \alpha \cr \gamma & \alpha & \beta \cr } } \right| = 0$$
$$ \Rightarrow $$ $${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0$$
$$ \Rightarrow $$ $$(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0$$
$$ \Rightarrow $$ $$( - a)[{a^2} - 3b] = 0$$
$$ \Rightarrow $$ $${a^2} = 3b$$ ($$ \because $$ a $$ \ne $$ 0)
$$ \Rightarrow {{{a^2}} \over b} = 3$$
Roots are $$\alpha$$, $$\beta$$, $$\gamma$$.
For non-trivial solutions,
$$\left| {\matrix{ \alpha & \beta & \gamma \cr \beta & \gamma & \alpha \cr \gamma & \alpha & \beta \cr } } \right| = 0$$
$$ \Rightarrow $$ $${\alpha ^3} + {\beta ^3} + {\gamma ^3} - 3\alpha \beta \gamma = 0$$
$$ \Rightarrow $$ $$(\alpha + \beta + \gamma )\left[ {{{\left( {\alpha + \beta + \alpha } \right)}^2} - 3\left( {\sum {\alpha \beta } } \right)} \right] = 0$$
$$ \Rightarrow $$ $$( - a)[{a^2} - 3b] = 0$$
$$ \Rightarrow $$ $${a^2} = 3b$$ ($$ \because $$ a $$ \ne $$ 0)
$$ \Rightarrow {{{a^2}} \over b} = 3$$
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