If α, β, γ are the roots of the equation x^3+p x^2+q x+r=0 , then (α+β)(β+γ)(γ+α)=

Correct answer: (B) r-p q. We have, x^3+p x^2+q x+r=0 α, β, γ are roots of equation ∴ α+β+γ =-p α β+β γ+α γ =q α β γ =-r Now, (α+β) β+γ)(γ+α) .=(α+β) β γ+α β+α γ+γ^2 ) =α β γ+α^2 β+α^2 γ+α γ^2+β^2 γ+α β^2+α β γ+β γ^2 =2 α β γ+α^2 β+α^2 γ+α γ^2+β^2 γ+α β^2+β γ^2 =(α+β+γ)(α β+β γ+α γ)-α β γ =-p(q)-(-r)=r-p q

Q54 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$, then $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=$

A
$p-q r$
Correct answer
B
$r-p q$
C
$q-r p$
D
$r+p q$

Solution

We have, $$ x^3+p x^2+q x+r=0 $$ $\alpha, \beta, \gamma$ are roots of equation $$ \begin{aligned} \therefore \quad \alpha+\beta+\gamma & =-p \\ \alpha \beta+\beta \gamma+\alpha \gamma & =q \\ \alpha \beta \gamma & =-r \end{aligned} $$ Now, $(\alpha+\beta) \beta+\gamma)(\gamma+\alpha)$ $$ \begin{aligned} & \left.=(\alpha+\beta) \beta \gamma+\alpha \beta+\alpha \gamma+\gamma^2\right) \\ & =\alpha \beta \gamma+\alpha^2 \beta+\alpha^2 \gamma+\alpha \gamma^2+\beta^2 \gamma+\alpha \beta^2+\alpha \beta \gamma+\beta \gamma^2 \\ & =2 \alpha \beta \gamma+\alpha^2 \beta+\alpha^2 \gamma+\alpha \gamma^2+\beta^2 \gamma+\alpha \beta^2+\beta \gamma^2 \\ & =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\alpha \gamma)-\alpha \beta \gamma \\ & =-p(q)-(-r)=r-p q \end{aligned} $$

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

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$, then $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=$

Solution

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