A) \[ac>0\]
B) \[a{{k}^{2}}+bk+c=0\]
C) \[ac<0\]
D) \[{{a}^{2}}{{k}^{2}}+abk+ac<0\]
Correct Answer: D
Solution :
Here \[a{{x}^{2}}+bx+c=a(x-\alpha )(x-\beta )\] Since \[\alpha ,\beta \] be the roots of\[a{{x}^{2}}+bx+c=0\]. Also \[\alpha <k<\beta ,\]so \[a(k-\alpha )(k-\beta )<0\] Also \[{{a}^{2}}{{k}^{2}}+abk+ac=a(a{{k}^{2}}+bk+c)\] \[={{a}^{2}}(k-\alpha )(k-\beta )<0\]Þ \[{{a}^{2}}{{k}^{2}}+abk+ac<0\]You need to login to perform this action.
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