A) \[-4\]
B) 1
C) 4
D) \[-\,2\]
E) 2
Correct Answer: D
Solution :
Since,\[\alpha \]and\[\beta \]are the roots of the equation \[a{{x}^{2}}+bx+c=0\] \[\therefore \] \[\alpha +\beta =-\frac{b}{a}\] and \[\alpha \beta =\frac{c}{a}\] But \[\alpha \beta =3\] \[\therefore \] \[3=\frac{c}{a}\Rightarrow c=3a\] ...(i) Also a, b, c are in AP. \[\therefore \] \[b=\frac{a+c}{2}\] \[\Rightarrow \] \[b=\frac{a+3a}{2}=2a\] Hence, \[\alpha +\beta =-\frac{b}{a}=-\frac{2a}{a}=-2\]You need to login to perform this action.
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