A) \[{{x}^{2}}+20x+16=0\]
B) \[{{x}^{2}}-10x+16=0\]
C) \[{{x}^{2}}+10x+16=0\]
D) \[{{x}^{2}}-10x-16=0\]
E) \[{{x}^{2}}+20x+32=0\]
Correct Answer: B
Solution :
Let\[\alpha \]and\[\beta \]be the roots of equation. Then, \[\alpha +\beta =10,\text{ }\alpha \text{ }\beta =16\] \[\therefore \]Required equation is \[{{x}^{2}}-(\alpha +\beta )x+\alpha \text{ }\beta =0\] \[\Rightarrow \] \[{{x}^{2}}-10x+16=0\]
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