A) \[{{x}^{2}}+18x-16=0\]
B) \[{{x}^{2}}-18x+16=0\]
C) \[{{x}^{2}}+18x+16=0\]
D) \[{{x}^{2}}-18x-16=0\]
Correct Answer: B
Solution :
Let the two numbers be\[\alpha \]and\[\beta ,\]then \[\frac{\alpha +\beta }{2}=9\]and\[\alpha \beta ={{4}^{2}}=16\] \[\therefore \]Required equation is\[{{x}^{2}}-(\alpha +\beta )+\alpha \beta =0\] \[\therefore \] \[{{x}^{2}}-18x+16=0\]
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