One root of the equation x2+6x+q=0 is twice the other root. find q and the roots.

Suppose one of the roots is a the second is 2a. The factors will be:

(x-a) and (x-2a)

thus

(x-a) (x-2a) = x^2+6x+q

x^2-xa-2ax+2a^2=x^2+6x+q

taking corresponding terms we shall have:

-2ax=6x

solving for a we get

-2a=6

thus

a=-3

but

q=2a^2

substituting the value of a we get:

q=2 (-3) ^2

q=2*9

q=18