
equations
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 (x2)(x4)=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?

For one root of $\displaystyle ax^2 + bx + c $ to be double the other
then
$\displaystyle \frac {b \sqrt{b^24ac} }{2a} = 2 \times \frac {b+ \sqrt{b^24ac} }{2a} $
solving above equation we get
$\displaystyle 2b^2=9ac$

Another solution...
$\displaystyle ax^2+bx+c$
if $\displaystyle x_1$ and $\displaystyle x_2$ are two roots, we demand on $\displaystyle x_2$ to be $\displaystyle x_2=2x_1$
Now, sum of the two roots is $\displaystyle (*) x_1+x_2=3x_1=\frac{b}{a}$ and their product is $\displaystyle x_1x_2=2x_1^2=\frac{c}{a}$
So, $\displaystyle x_1=\sqrt{\frac{c}{2a}}$ put this to $\displaystyle (*)$, and you get what you need to prove.