Question

(1)/(4p)(x-h)^(2)+k=0

Answer 1

Assume that $p\ \textgreater \ 0$ and multiply the equation $\frac{1}{4p} (x-h)^2+k=0$ by 4p. Then you obtain the equation $(x-h)^2+4pk=0$.

1) If k>0, then 4pk>0 and the equation doesn't have solutions, because $(x-h)^2=-4pk$ and this is unreal.

2) If k=0, then 4pk=0 and $(x-h)^2=0$. There is one solution x=h.

2) If k<0, then 4pk<0 and the equation
$(x-h)^2=-4pk>0$ has two different solutions $x_1=h+ \sqrt{-4pk}$ and $x_2=h- \sqrt{-4hk}$.