A) \[{{x}^{2}}+x+1=0\]
B) \[{{x}^{2}}-x+1=0\]
C) \[{{x}^{2}}-x-1=0\]
D) \[{{x}^{2}}+x-1=0\]
Correct Answer: A
Solution :
Given a and p are roots of \[{{x}^{2}}+x+1=0\] \[\Rightarrow \] \[\alpha +\beta =-1\] and \[\alpha \beta =1\] Now, \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{(\alpha +{{\beta }^{2}})-2\alpha \beta }{\alpha \beta }\] \[=\frac{1-2}{1}=-1\] \[\frac{\alpha }{\beta }.\frac{\beta }{\alpha }=1\] \[\therefore \]Equation having roots \[\frac{\alpha }{\beta }\]and\[\frac{\beta }{\alpha }\]is \[{{x}^{2}}+x+1=0\]
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