a. the equation 3x^2 + 9x=17+6kx have roots that have equal magnitude but opposite in signs.
b. the graph of y=x^2+kx+k+8 intersects the x axis at two distinct points.
Use the quadratic formula.
(a)
$\displaystyle 3x^2+9x=17+6kx\Rightarrow\ 3x^2+9x-6kx-17=0$
$\displaystyle \displaystyle\ ax^2+bx+c=0\Rightarrow\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle 3x^2+(9-6k)x+(-17)=0$
For part (a), you require $\displaystyle \pm(value)$, so the coefficient $\displaystyle b$ needs to be zero.
(b)
$\displaystyle x^2+kx+(k+8)=0$
The two points will be the same if $\displaystyle b^2-4ac=0$
so you need to discover how $\displaystyle b^2-4ac\ \ne\ 0$