A) \[4ac<{{b}^{2}}\]
B) \[4ac>{{b}^{2}}\]
C) \[ac<{{b}^{2}}\]
D) \[ac>{{b}^{2}}\]
Correct Answer: B
Solution :
Let\[f(x)=a{{x}^{2}}+bx+c\]. Then\[f(0)=c\]. Thus the graph of \[y=f(x)\]meets y-axis at (0, c). If\[c>0\], then by hypothesis \[f(x)>0\] This means that the curve \[y=f(x)\] does not meet x-axis. If\[c<0\], then by hypothesis\[f(x)<0\], which means that the curve \[y=f(x)\] is always below x-axis and so it does not intersect with x-axis. Thus in both cases \[y=f(x)\] does not intersect with x-axis i.e. \[f(x)\ne 0\]for any real x. Hence \[f(x)=0\]i.e. \[a{{x}^{2}}+bx+c=0\] has imaginary roots and so\[{{b}^{2}}<4ac\].
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