Find $K$ in the equation $ 24x^3-14x^2-63x+K=0$ if one root is twice more the another. I am given the equation $ 24x^3-14x^2-63x+K=0$  and I am asked to find the values for $K$ If one root is double the other root.
How would I solve this?
I am trying to solve this by taking two roots as A,2A and the third B then by using sum and products of roots I wrote three equations.
 A: By using Viete's formulae for roots $x$, $2x$, $y$ of equation $24x^3-14x^2-63x+K=0$, we get system of equations $$3x+y=\frac{7}{12},$$ $$3xy+2x^2=-\frac{21}{8},$$ $$K=-48x^2y.$$
From the first equation $y=\frac{7}{12}-3x$, substitute this into second equation in order to get $8x^2-2x-3=0$.  Now it is easy to conclude that solutions are $$(x,y)\in\{(\frac{3}{4},-\frac{5}{3}),(-\frac{1}{2},\frac{25}{12})\}.$$
Finally, calculate $K$ from the third equation, for those values $x$ and $y$:
$$K\in\{45,-25\}.$$
A: The solution is computable with simple mental arithmetic if we  eliminate fractions by scaling the polynomial by $72$ to transform it to have leading coeff $=1\ $ (AC method). $ $ This yields
$$\quad\ X^3 - 7 X^2 -63\cdot 6 X + 72K = 0,\ \ X = 12x$$
Consider more generally $\ X^3 - a\, X^2 + b\, X + c = 0\ $ with roots $\,X,\,2X,\, Y.\ $ By Vieta
$$3X+Y = a,\ \ 3XY+2X^2 = b$$
Eliminate $Y$ and scale by $\,7\,$ to get  $\ (7X)^2 - 3a (7X) + 7b = 0\ $ with solution
$$7X = (3a\pm d)/2,\ \ d = \sqrt{9a^2-28b}\qquad $$
In OP $\,\ 9a^2\!-28b = 9(7)^2\!+4\ 7\ (3^2\ 7\ 6) = 3^2 7^2 (1\!+\!24)   =(3\ 7 \ 5)^2 = 105^2 $
Therefore $\ X = (21\pm 105)/14 = 9,\, -6,\ \ $ so $\ \ Y = 7-3X = -20,\, 25$ 
which $ $  implies  $\ \ x = X/12 = 3/4,\, -1/2\ \ $ and $\ \ y = Y/12 = -5/3,\, 25/12$
A: Suppose $x^3-\frac{7}{12}x^2-\frac{63}{24}x+\frac{K}{24}=(x-a)(x-2a)(x-b)$.
So $\frac{63}{24}=2a^2+3ab$ and $\frac{7}{12}=-3a-b$.
We conclude $b=-3a-\frac{7}{12}$.
So we must solve $\frac{63}{24}=2a^2-3a(3a+\frac{7}{12})$.
From here you can find $a$, and once you have $a$ you get $b$.
Finally $K=24(-2a^2b)$
