Find the value of a for which one root of the quadratic equation (a2-5a+3)x2+(3a-1)x+2=0 is twice as large as the other.How can we eliminate alpha here? Please solve the problem?
Let one root be T then the other root will be 2T.
Comparing equation (a² - 5a + 3)x² + (3a - 1)x + 2 = 0 with Ax² + Bx + C =0
We get A = (a² - 5a + 3), B = (3a - 1) , C = 2
Sum of roots =
3T = (1)
And product of roots = 2T2 = (2)
Comparing (1) and (2) , we have
Hence if we put a = 2/3
Then the roots will be -3 and -6 .