If α, β, γ are the roots of the equation x^3+a x^2+b x+c=0 , then α^-1+β^-1+γ^-1 is equal to

Correct answer: (C) -b/c. Let α, β, γ are roots of equation x^3+a x^2+b x+c=0 α+β+γ=-a ∴ α β+β γ+γ α=b Now, α^-1+β^-1+γ^-1=1/(α)+1/(β)+1/(γ) =(α β+β γ+γ α)/(α β γ)=b/(-c)=-b/c

Q131 / 200 · TS EAMCET 2002 · Mathematics

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2+b x+c=0$, then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

A
$\frac{a}{c}$
B
$\frac{c}{a}$
Correct answer
C
$-\frac{b}{c}$
D
$\frac{b}{a}$

Solution

Let $\alpha, \beta, \gamma$ are roots of equation $$ \begin{aligned} & x^3+a x^2+b x+c=0 \\ & \alpha+\beta+\gamma=-a \\ & \therefore \quad \alpha \beta+\beta \gamma+\gamma \alpha=b \\ & \text { Now, } \alpha^{-1}+\beta^{-1}+\gamma^{-1}=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} \\ &=\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}=\frac{b}{-c}=-\frac{b}{c} \end{aligned} $$

Hence, the correct answer is option: 'C'.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2+b x+c=0$, then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

Solution

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