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

Let α and 2αbe the roots of (1), thena2−5a+3α2+(3a−1)α+2=0                   (2)and a2−5a+34α2+(3a−1)(2α)+2=0    (3)Multiplying (2) by 4 and subtracting it form (3) we get(3a−1)(2α)+6=0Clearly a≠1/3. Therefore, α=−3/(3a−1)Putting this value in (2) we get a2−5a+3(9)−(3a−1)2(3)+2(3a−1)2=0⇒ 9a2−45a+27−9a2−6a+1=0⇒  −39a+26=0⇒ a=2/3.For a=2/3, the equation becomes x2+9x+18=0, whose roots are – 3, – 6.