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\[\alpha \]and\[\beta \]be two numbers whose arithmetic mean is 9 and geometric mean is 4. \[\therefore \] \[\alpha +\beta =18\] ...(i) and \[\alpha \beta =16\] ...(ii) \[\therefore \]Required equation is \[{{x}^{2}}-(\alpha +\beta )x+(\alpha \beta )=0\] \[\Rightarrow \] \[{{x}^{2}}-18x+16=0\] [using Eqs. (i) and (ii)]You need to login to perform this action.
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