Question

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

Multiply the equation by 4p. Explain how different values of k affect the number of zeros of the polynomial. Consider k > 0, k = 0, and k < 0. Assume p > 0.

Answer 1

Assume that
`p\ \textgreater \ 0` and multiply the equation
`(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<0` 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+ √(-4pk)` and
`x_2=h- √(-4hk)`.