In the equation `x^(2)+kx+54=0`, one root is twice the other root. The value(s) of k is (are)

Options

A. `-5.2`
B. `+-15.6`
C. `22.0`
D. `+-5.2`

Explanation

If the roots are r and 2r, their `sum=-(b)/(a)=3r=-(k)/(1)` and their product `=(c)/(a)=` ltbr. `2r^(2)=(57)/(1)`. Therefore, `r=+-sqrt(27) and k=+-sqrt(27)=+-15.6`
Alternative solution: if the roots are r and 2r, `(x-r)(x-2r)=0`. Multiply to obtain `x^(2)-3r+2r^(2)=0`, which represents `x^(2)+kx+54=0`. thus, `-3r=k and 2r^(2)=54`.
since `r=-(k)/(3),` then `2(-(k)/(3))=54` and `k=+-3sqrt(27)~~+-15.6`