A) \[4-\frac{2b}{a}+\frac{c}{a}<0\]
B) \[4+\frac{2b}{a}-\frac{c}{a}<0\]
C) \[4-\frac{2b}{a}+\frac{c}{a}=0\]
D) \[4+\frac{2b}{a}+\frac{c}{a}=0\]
Correct Answer: A
Solution :
Given \[a,\,\,b,\,\,c\] are real, \[a{{x}^{2}}+bx+c=0\] has two real roots \[\alpha \] and \[\beta \] where \[\alpha <-2\] and \[\beta >2\] \[\Rightarrow \] \[f(-2)<0\]and\[f(2)>0\] \[\Rightarrow \] \[4a-2b+c<0\] \[\Rightarrow \]\[4-\frac{2b}{a}+\frac{c}{a}<0\]
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