A) \[{{x}^{2}}+x-6=0\]
B) \[{{x}^{2}}-x+6=0\]
C) \[6{{x}^{2}}+x+1=0\]
D) \[{{x}^{2}}-6x+1=0\]
Correct Answer: A
Solution :
Let roots are \[\alpha \] and\[\beta \] . Given \[\alpha +\beta =-1\] \[\frac{1}{\alpha }+\frac{1}{\beta }=\frac{1}{6}\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=\frac{1}{6}\Rightarrow \alpha \beta =-6\] Hence the equation, \[{{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0\] Þ \[{{x}^{2}}+x-6=0\]
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