A) 2/3
B) -2/3
C) 1/3
D) -1/3
Correct Answer: A
Solution :
If \[\alpha \] and \[2\alpha \] are the roots of the equation \[a{{x}^{2}}+bx+c=0\], then \[2{{b}^{2}}=9ac\]. Since, one root of the quadratic equation \[({{a}^{2}}-5a+3){{x}^{2}}+(3a-1)x+2=0\] is twice as large as the other, then \[2{{(3a-1)}^{2}}=2\times 9\,({{a}^{2}}-5a+3)\] \[\Rightarrow \] \[9{{a}^{2}}-6a+1=9{{a}^{2}}-45a+27\] \[\Rightarrow \] \[45a-6a=27-1\] \[\Rightarrow \] \[a=\frac{26}{39}=\frac{2}{3}\] Alternate Solution The given equation is \[({{a}^{2}}-5a+3){{x}^{2}}+(3a-1)x-2=0\] Let a and 2a be the roots of this equation, then \[\alpha +2\alpha =-\frac{(3a-1)}{({{a}^{2}}-5a+3)}\] \[\Rightarrow \] \[3\alpha =-\frac{(3a-1)}{({{a}^{2}}-5a+3)}\] and \[\alpha \,.\,\,2\alpha =\frac{2}{({{a}^{2}}-5a+3)}\] \[\Rightarrow \] \[2{{\alpha }^{2}}=\frac{2}{({{a}^{2}}-5a+3)}\] \[\Rightarrow 2{{\left[ \frac{-(3a-1)}{3\,({{a}^{2}}-5a+3)} \right]}^{2}}=\frac{2}{({{a}^{2}}-5a+3)}\] \[\Rightarrow \] \[\frac{{{(3a-1)}^{2}}}{9\,{{({{a}^{2}}-5a+3)}^{2}}}=\frac{1}{({{a}^{2}}-5a+3)}\] \[\Rightarrow \] \[{{(3a-1)}^{2}}=9\,({{a}^{2}}-5a+3)\] \[\Rightarrow \] \[9{{a}^{2}}+1-6a=9{{a}^{2}}-45a+27\] \[\Rightarrow \] \[45a-6a\,=27\,-1\Rightarrow \,\,39a=26\] \[\therefore \] \[a=\frac{2}{3}\]
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