If one root of the equation $$ax^{2} + bx + c = 0$$ is double of the other, then $$2b^{2}$$ =

The given equation is $$ax^{2} + bx + c = 0$$
Roots of the given quadratic equation are $$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
We know that one of the 2 roots is double the other. 
Therefore, $$\dfrac{-b+\sqrt{b^2-4ac}}{2a}$$ =$$2*\dfrac{-b-\sqrt{b^2-4ac}}{2a}$$
=> $$-b +\sqrt{b^2-4ac} = -2b -2\sqrt{b^2-4ac}$$
=> $$b = -3\sqrt{b^2-4ac}$$
Squaring on both sides, we get, 
$$b^2 = 9*(b^2-4ac)$$
$$8b^2 = 36 ac$$
$$2b^2 = 9ac$$.
Therefore, option A is the right answer. 

Alternately

Suppose two roots are x, 2x

Sum of the roots= 3x=-b/a

Product of the roots=$$2x^{2}$$ = c/a

Putting the value of x from the first eqn.

 We get $$2b^2 = 9ac$$..