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\[\beta \]are two numbers whose arithmetic mean is 9 and geometric mean is 4. \[\therefore \] \[\alpha +\beta =18,\alpha \beta =16\] Required equation is \[{{x}^{2}}-(\alpha -\beta )x+\alpha \beta =0\] \[\Rightarrow \] \[{{x}^{2}}-18x+16=0\]
You need to login to perform this action.
You will be redirected in
3 sec
