The value of '$$a$$' for which one root of the quadratic equation
$$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$
is twice as large as the other is
A
$$ - {1 \over 3}$$
B
$$ {2 \over 3}$$
C
$$ - {2 \over 3}$$
D
$$ {1 \over 3}$$
Explanation
Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then
If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in