writing and solving quadratic functions

Question

A movie theatre models the revenue from ticket sales in one day as a function of the ticket price, p. Here are two expressions defining the same revenue function.

p(120−4p)�(120−4�)

120p−4p2120�−4�2

According to this model, how high would the ticket price have to be for the theater to make $0 in revenue? Explain your reasoning.

Responses

1. $45; When solving, p=0 �=0  and 120−4p=0120−4�=0. Therefore, p=0 �=0  or p=45�=45. Tickets cannot equal $0, so therefore each ticket costs $45.
2. $45; When solving, p=0 �=0  and 120 minus 4 p is equal to 0. Therefore, p=0 �=0  or p is equal to 45. Tickets cannot equal $0, so therefore each ticket costs $45.
3. $45; When solving, p=0 �=0  and 120−4p=0120−4�=0. Therefore, p=0 �=0  and p=45�=45. Tickets are equal to $0 and $45.
4. $45; When solving, p=0 �=0  and 120 minus 4 p is equal to 0. Therefore, p=0 �=0  and p is equal to 45. Tickets are equal to $0 and $45.
5. $30; When solving, p=0 �=0  and 120−4p=0120−4�=0. Therefore, p=0 �=0  or p=30�=30. Tickets cannot equal $0, so therefore each ticket costs $30.
6. $30; When solving, p=0 �=0  and 120 minus 4 p is equal to 0. Therefore, p=0 �=0  or p is equal to 30. Tickets cannot equal $0, so therefore each ticket costs $30.
7. $0 and $30; When solving, p=0 �=0  and 120−4p=0120−4�=0. Therefore, p=0 �=0  and p=30�=30. Tickets are equal to $0 and $30.

Answer 1