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(inu,v,w) given by \(\alpha \mathrm{u}+\beta \mathrm{v}+\gamma \mathrm{w}=0 ; \beta \mathrm{u}+\gamma \mathrm{v}+\alpha \mathrm{w}=0 ; \gamma \mathrm{u}+\alpha \mathrm{v}+\beta \mathrm{w}=0\) has non-trivial solution, then the value of \(\frac{\mathrm{a}^{2}}{\mathrm{~b}}\) is :

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Determinants

Let

α, β, γ

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(inu,v,w) given by

α u + β v + γ w = 0; β u + γ v + α w = 0; γ u + α v + β w = 0

has non-trivial solution, then the value of

\frac{a ^{2}}{b}

is :

Answer: (B)

A homogeneous system has a non-trivial solution only when its coefficient determinant vanishes, so

α β γ β γ α γ α β = 0

Expanding this circulant determinant gives

- (α + β + γ) (α^{2} + β^{2} + γ^{2} - α β - β γ - γ α) = 0

. Since

α + β + γ = - a

and

a \neq = 0

, the first factor cannot vanish, so

\frac{1}{2} [(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2}] = 0

The roots are real, so each square is non-negative and all three must vanish, giving

α = β = γ

. Then

a = - 3 α

and

b = 3 α^{2}

, with

α \neq = 0

because

b \neq = 0

, so

\frac{a ^{2}}{b} = \frac{9 α ^{2}}{3 α ^{2}} = 3

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option