The value of ‘a’ for which one root of the quadratic equation (a² − 5a + 3)x² + (3a − 1)x + 2 = 0 is twice the other, is:

Correct option is D

Given:

Equation: (a² − 5a + 3)x² + (3a − 1)x + 2 = 0
Let roots be α and 2α.

Concept used:
For quadratic equation Ax² + Bx + C = 0,
Sum of roots = −B/A and Product of roots = C/A

Formula used:
α + 2α = −B/A and α(2α) = C/A

Solution:
A = (a² − 5a + 3), B = (3a − 1), C = 2

Using sum of roots:

​$$\alpha + 2\alpha = -\frac{B}{A} \Rightarrow 3\alpha = -\frac{3a - 1}{a^2 - 5a + 3}$$​​

​$$\Rightarrow \alpha = -\frac{3a - 1}{3(a^2 - 5a + 3)}$$​​

Using product of roots:
$$\alpha \cdot 2\alpha = \frac{C}{A} \Rightarrow 2\alpha^2 = \frac{2}{a^2 - 5a + 3} \\\Rightarrow \alpha^2 = \frac{1}{a^2 - 5a + 3}$$​​

Substitute into 

​$$\left( \frac{3a - 1}{3(a^2 - 5a + 3)} \right)^2 = \frac{1}{a^2 - 5a + 3} \\\Rightarrow \frac{(3a - 1)^2}{9(a^2 - 5a + 3)^2} = \frac{1}{a^2 - 5a + 3}$$​​

Simplify:
(3a − 1)² = 9(a² − 5a + 3)
=> 9a² − 6a + 1 = 9a² − 45a + 27
=> −6a + 1 = −45a + 27
=> 39a = 26
​$$\Rightarrow a = \frac{26}{39} = \frac{2}{3}$$​​

Correct answer is (d) 2/3.