A) \[2<a<3\]
B) \[a>3\]
C) \[-3<a<3\]
D) \[a<-2\]
Correct Answer: D
Solution :
If the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] exceed a number k, then \[a{{k}^{2}}+bk+c>0\] if \[a>0,\] \[{{b}^{2}}-4ac\ge 0\] and sum of the roots \[>2k\] Therefore, if the roots of \[{{x}^{2}}+x+a=0\] exceed a number a, then \[{{a}^{2}}+a+a>0,1-4a\ge 0\] and \[-1>2a\] Þ \[a(a+2)>0,\]\[a\le \frac{1}{4}\]and \[a<-\frac{1}{2}\] Þ \[a>0\,\text{or}\,a<-2,a<\frac{1}{4}\]and \[a<-\frac{1}{2}\] Hence\[a<-2\].
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