Let α and β be the roots of x3 + ax2 + bx+ c = 0 satisfying the relation αβ + 1 = 0. then c2 + ac + b + 1 =

Let the third root be γ. Thus, α+β+γ=−α,αβ+αγ+βγ=b,αβγ=−c Given condition is αβ+1=0⇒ (−1)γ=−c⇒ γ=cAlso, αβ+αγ+βγ=b ⇒ αβ+(α+β)γ=b⇒ −1+γ(−α−γ)=b⇒ 1+γ(α+γ)=−b⇒ γ2+aγ+b+1=0⇒ c2+ac+b+1=0Hence, the result.

Let α and β be the roots of x3 + ax2 + bx+ c = 0 satisfying the relation αβ + 1 = 0. then c2 + ac + b + 1 =

Let the third root be γ. Thus, α+β+γ=−α,αβ+αγ+βγ=b,αβγ=−c Given condition is αβ+1=0⇒ (−1)γ=−c⇒ γ=cAlso, αβ+αγ+βγ=b ⇒ αβ+(α+β)γ=b⇒ −1+γ(−α−γ)=b⇒ 1+γ(α+γ)=−b⇒ γ2+aγ+b+1=0⇒ c2+ac+b+1=0Hence, the result.

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Let $α$ and $β$ be the roots of x³ + ax² + bx+ c = 0 satisfying the relation $α β$ + 1 = 0. then c² + ac + b + 1 =

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$2 c^{2} + a c + b + 1 = 0$

$c^{2} - a c + b - 1 = 0$

zero

$c^{2} + a c + b + 1 = 0$

answer is D.

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Detailed Solution

Let the third root be $γ$ . Thus, $α + β + γ = - α, α β + α γ + β γ = b, α β γ = - c$ Given condition is $α β + 1 = 0$

$\begin{array}{l} \Rightarrow (- 1) γ = - c \\ \Rightarrow γ = c \end{array}$

Also,

$α β + α γ + β γ = b \begin{array}{ll} \Rightarrow α β + (α + β) γ = b \\ \Rightarrow - 1 + γ (- α - γ) = b \\ \Rightarrow 1 + γ (α + γ) = - b \\ \Rightarrow γ^{2} + a γ + b + 1 = 0 \\ \Rightarrow c^{2} + a c + b + 1 = 0 \end{array}$