For one root of to be double the other
solving above equation we get
For one root of ax(squared) + bx + c to be double the other, the coefficients a, b, c, must be related as follows:
a. 4b(squared) = 9c
b. 2b(squared) = 9ac
c. 2b(squared) = 9a
d. b(squared) - 8ac = 0
e. 9b(squared) = 2ac
I got the answer b by using (x-2)(x-4)=0.
I plugged in a=1 b= -6 c=8 but I think the problem wasn't meant to be solved that way. Is there a better way to do it?