• # question_answer The value of 'a' for which one root of the quadratic equation $({{a}^{2}}-5a+3){{x}^{2}}+(3a-1)x+2=0$ is twice as large as the other, is [AIEEE 2003] A) $\frac{2}{3}$B) $-\frac{2}{3}$C) $\frac{1}{3}$D) $-\frac{1}{3}$

Let the roots are a and 2a Þ  $\alpha +2\alpha =\frac{1-3a}{{{a}^{2}}-5a+3}$  and $\,\alpha .2\alpha =\frac{2}{{{a}^{2}}-5a+3}$ Þ $2\left[ \frac{1}{9}\frac{{{(1-3a)}^{2}}}{{{({{a}^{2}}-5a+3)}^{2}}} \right]=\frac{2}{{{a}^{2}}-5a+3}$ Þ  $\frac{{{(1-3a)}^{2}}}{({{a}^{2}}-5a+3)}=9$$\Rightarrow 9{{a}^{2}}-6a+1=9{{a}^{2}}-45a+27$ Þ $39a=26$$\Rightarrow a=\frac{2}{3}$.