Answer:
Option C
Explanation:
Given , $\alpha$ and $\beta$ are roots of $x^{2}+2x+c=0$
$\therefore$ Sum of roots , $\alpha+\beta$= -coefficient of x / coefficient of x2
$\Rightarrow$ $\alpha+\beta=\frac{-2}{1}=-2$ .......(i)
and products of roots , $\alpha \beta$= costant term/ coefficient of x2
$\Rightarrow$ $\alpha \beta =\frac{c}{1}=c$ .......(ii)
Since , $\alpha^{2}+\beta^{2}=4$ [given]
$\Rightarrow$ $(\alpha+\beta)(\alpha^{2}+\beta^{2}-\alpha \beta)=4$
$\left[\because a^{3}+b^{3}=(a+b)(a^{2}+b^{2}-ab)\right]$
$\Rightarrow$ $(\alpha+\beta)[(\alpha^{2}+\beta^{2}-3\alpha\beta)]=4$
$\left[\because a^{2}+b^{2}=(a+b)^{2}-2\alpha\beta)\right]$
$\Rightarrow$ $ (-2)[(-2)^{2}-3 \times c]=4$ [ From Eqs,(i) and (ii)]
$\Rightarrow$ (-2)[4-3c]=4
$\Rightarrow$ $4-3c=\frac{-4}{2}$
$\Rightarrow$ $4-3c=-2$
$\Rightarrow$ $-3c=-2-4$
$\Rightarrow$ -3c=-6
c=2
