Q86 / 160 · TS EAMCET 2017 · Mathematics

$\alpha$ and $\beta$ are the roots of $x^2+2 x+c=0$. If $\alpha^3+\beta^3=4$, then the value of $c$ is

  • A
    -2
  • B
    3
  • Correct answer
    C
    2
  • D
    4

Solution

Given, $\alpha$ and $\beta$ are the roots of $x^2+2 x+c=0$. $\therefore$ Sum of roots, $\alpha+\beta=\frac{- \text { coefficient of } x}{\text { coefficient of } x^2}$ $\Rightarrow \quad \alpha+\beta=\frac{-2}{1}=-2$ and Product of roots, $\alpha \beta=\frac{\text { constant term }}{\text { coefficient of } x^2}$ $\Rightarrow \quad \alpha \beta=\frac{c}{1}=c$ Since, $\quad \alpha^3+\beta^3=4$ [given] $$ \begin{aligned} & \Rightarrow(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)=4 \\ & {\left[\because a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right]} \\ & \Rightarrow(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]=4 \\ & {\left[\because a^2+b^2=(a+b)^2-2 \alpha \beta\right]} \\ & \Rightarrow(-2)\left[(-2)^2-3 \times c\right]=4 \quad \text { [From Eqs. (i) and (ii)] } \\ & \Rightarrow \quad(-2)[4-3 c]=4 \\ & \Rightarrow \quad 4-3 c=\frac{-4}{2} \\ & \Rightarrow \quad 4-3 c=-2 \\ & \Rightarrow \quad-3 c=-2-4 \\ & \Rightarrow \quad-3 c=-6 \\ & c=2 \\ & \end{aligned} $$

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

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