I see, I did this using Vieta's Formula and got the same answer. Thank you!!!
I can do it a third way, although to be honest I do not know what Vieta's formula is
Sum of the roots in any polynomial is = -b/a
In this case that is -12/1 = -12
let one root be T the other is twice the size so
T+2T=-12
3T=-12
T= -4
So the roots are -4 and -8 so k is 32
One root of \(x^2+12x+k=0\) is twice the other root.
Find \(k\).
Let the roots are \(x_1\), \(x_2\).
\(\begin{array}{|rcll|} \hline x^2+12x+k=0 &=& (x-x_1)(x-x_2)= 0 \quad | \quad x_2 = 2x_1 \\\\ &=& (x-x_1)\Big(x-(2x_1)\Big) \\ &=& (x-x_1)(x-2x_1) \\ &=& x^2-2x_1x-x_1x+2x^2_1 \\ x^2+12x+k=0 &=& x^2-3x_1x+2x^2_1 \\ && \text{compare} \\ && \boxed{12=-3x_1\ \text{ or }\ x_1=-4 \\ k= 2x^2_1\ \text{ or }\ k=2*(-4)^2=32 } \\ \hline \end{array} \)
\(\boxed{k=32}\)