Originally Posted by
Mush $\displaystyle f(x+k) = \frac{-1}{2}(x+k)^2 -5(x+k) + 4 $
$\displaystyle f(x+k) = \frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4 $
$\displaystyle f(x+k) = \frac{-1}{2}x^2-\frac{1}{2}2kx-\frac{1}{2}k^2 -5x-5k + 4 $
$\displaystyle f(x+k) = \frac{-1}{2}x^2+x(-k-5) - \frac{k^2}{2} -5k + 4$
The discriminant must be zero for equal roots:
** $\displaystyle b^2-4ac = 0 $
** $\displaystyle a = \frac{-1}{2}$
** $\displaystyle b = (-k-5)$
** $\displaystyle c = - \frac{k^2}{2} -5k + 4$
Insert these into the equation. Expand. Then solve for k.