Given equation
(a2−5a+3)x2−3(a−1)x+2=0
For quadratic
a2−5a+3≠0⇒a≠5±√132
Let the roots of the equation m,n
Now m=2n
So,
m+n=3(a−1)a2−5a+3⇒n=a−1a2−5a+3⋯(1)
and
2n2=2a2−5a+3⇒n2=1a2−5a+3⋯(2)
From (1) and (2)
(a−1a2−5a+3)2=1a2−5a+3⇒(a−1)2=(a2−5a+3)⇒a2+1−2a=a2−5a+3∴3a=2