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

Answer:

Step-by-step 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

Answer:

Given equation:

• $\frac{1}{4p} (x - h)^2 + k = 0$

Multiply this by 4p to get:

• (x - h)² + 4pk = 0

Rewrite this as:

• (x - h)² = 4pk

1) If p > 0 and k < 0

• (x - h)² < 0, then no real solutions as left side is always positive

2) If p > 0 and k = 0

• (x - h)² = 0, then x = h, one solution = 1 zero

3) If p > 0 and k > 0

• (x - h)² > 0, then there 2 real solutions = 2 zeros:
• x = h ± 2$\sqrt{pk}$