A) \[2x-3\]
B) \[4x-7\]
C) \[{{x}^{2}}-3x+1\]
D) \[{{x}^{2}}-6x+9\]
Correct Answer: C
Solution :
[c] Let the equation \[{{x}^{2}}-6x+a=0\] have roots \[\alpha \]and \[\beta \]. Also, let the equation \[{{x}^{2}}-ex+6=0\]have roots \[\alpha \]and \[\beta \]. Given that \[\frac{\beta }{\gamma }=\frac{4}{3}\] \[\Rightarrow \,\,\,\frac{\alpha \beta }{\alpha \gamma }=\frac{4}{3}\] \[\Rightarrow \,\,\,\frac{a}{6}=\frac{4}{3}\] \[\Rightarrow \,\,\,a=8\] So, first equation is \[{{x}^{2}}-6x+8=0,\]having roots \[x=2,4\]. For, \[\alpha =2\] and \[\beta =4,\] \[\gamma =\frac{3\beta }{4}=3\] For \[\alpha =4\] and \[\beta =2,\]\[\gamma =\frac{3\beta }{4}=\frac{3}{2},\]which is not an integer. Hence, common root is\[\alpha =2.\]
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