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

Explanation:

Assume that and multiply the equation by 4p. Then you obtain the equation (x-h)^2+4pk=0.

1) If k>0, then 4pk>0 and the equation does not have real solutions and there is no zero.

2) If k=0, then 4pk=0 and . There is one solution x=h and there is one zero.

2) If k<0, then 4pk<0 and the equation has two different solutions and there are two zeros.

Answer 2

Explanation:

Given p>0, multiply the equation by 4p: (1/4p)*(x-h)^(2)+k=0

(x-h)^2+4kp = 0

k>0

4kp>0

(x-h)^2 = -4kp

So x has imaginary roots only. There is no real zeros of the polynomial.

k=0

4kp=0

(x-h)^2 = 0

x=h

So x has one real root and the polynomial has one zero

k<0

4kp<0

(x-h)^2 = -4kp

So x has two real roots and the polynomial has two real zeros.