Lesson 2

This warm-up activates what students know about interpreting equations in context and about solving for a variable. The given equation is linear and is relatively straightforward. The work prepares students to reason about quadratic equations in the lesson.

To find the unknown input in each question, students might:

• Try different values of \(t\) until they find one that yields the specified value of \(c\).
• Reason backwards (subtract 2.5 from 62.50 and then divide the result by 12) without writing out the steps.
• Solve \(62.5=12t+2.5\) and \(278.5=12t+2.5\) by performing the same operation to each side to isolate \(t\).

Ask students to share their responses and reasoning. Highlight the different strategies used to answer the questions.

If no students thought of the situation in terms of a function and wrote an equation, point out that we can think about cost as being a function of the number of tickets. In answering the questions, we were looking for the inputs that produce different outputs. This can be done by solving equations. In the case of linear equations, we can “do the same thing to each side” to isolate the variable.

This task prompts students to try different ways to solve a quadratic equation. They are familiar with solving equations by performing the same operation to each side of an equation, but here they see that this is not really a workable strategy. Students are also discouraged from using a graph to solve the equation. Because they do not yet know an efficient way to use algebra to solve \( \text-16t^2 + 80t + 64=0\), they need to try different strategies and persevere in problem solving (MP1).

As students work, notice those who use strategies listed in the Activity Synthesis. Ask them to share their approach during discussion.

Arrange students in groups of 2. Give them a few minutes of quiet think time and then time to collaborate on solving the equations.

No graphing technology should be used in this activity.

Poll the class on their solutions to the equation and record and display the solutions for all to see. Then, ask some students to share their strategies and any associated challenges. If not mentioned by students, discuss the limitations of these approaches:

• Isolating the variable: If we try to solve by performing the same operation to each side of the equation, we quickly get stuck.

\(\displaystyle \begin {align} \text-16t^2 + 80t + 64 &=0 &\qquad& \text {Original equation}\\\\ \text-16t^2 + 80t &= \text-64 &\qquad& \text {Subtract 64 from each side}\\\\ t^2-5t &=4 &\qquad& \text {Divide each side by -16}\\ \end {align}\)

But then what? If we add \(5t\) to each side, we now have a variable on both sides and cannot combine any like terms. We could multiply or divide each side by any constant we wish, but we are no closer to isolating \(t\).

• Guessing and checking: We can evaluate the quadratic expression at different values of \(t\) until the expression has a value that is 0 or close to 0. For example, when \(t\) is 4, the expression has a value of 128. At \(t=5\), it has a value of 64, and at \(t=6\), it has a value of -32. That means \(t\) is between 5 and 6, so we need to try different decimal values in that range.

  This process is laborious, and may not get us to a precise solution.

• Graphing: Students may suggest that a graph would allow them to solve the problem much more quickly. Use graphing technology to demonstrate that if we graph the equation \(y=f(t)\), an approximate solution given is 5.702, as shown in the image.

  

  

  If we evaluate \(f(5.702)\), however, we get roughly -0.044864, rather than exactly 0.

  A graph is useful for approximating values, but it isn’t always possible to use it to find exact values.

Tell students that in this unit they will learn some efficient strategies for solving equations like these.

This activity aims to show that it is relatively easy to solve a quadratic equation when one side of the equation is zero and the other side is a quadratic expression in factored form, and that it may be a little tricky to solve the equation otherwise.

The activity prompts students to recall what they learned in an earlier unit, that the zeros of a function correspond to the horizontal intercepts of the graph representing that function, and that the zeros are the solutions to an equation of the form \(\text{quadratic expression} = 0\).

As students make sense of the equations and ways to use them to solve a contextual problem, they practice reasoning quantitatively and abstractly (MP2).

Keep students in groups of 2.

Read the opening paragraph in the task statement. Give students a moment to think about how much the school would collect if they sell the tickets at $5 each (it would collect $875). Briefly survey how they found out the answer. (Students are likely to have used the factored form because it lends itself to simpler calculations.)

Ask students to recall the form in which each quadratic expression is written. Then, give them a minute to talk to a partner and recall at least two things about each form and what the form might tell us about the graph of the function that the expression defines.

Record their responses for all to see. If no students mentioned the connection between either of the forms to the horizontal intercepts of the graph or the zeros of the function, ask them about it.

Tell students that they do not need to find the ticket prices in the second question, but only to think about how to go about doing so.

If students struggle to connect the expressions that define the function to the questions, ask them what the input and output of the function represent. If students struggle with the first question, ask them what values of \(p\) would yield a value of 0 for the expression.

Invite students to share their responses and strategies. Make sure students see that the first question can be represented by solving the equation \(p(200-5p)=0\), and that the second question can be represented by solving either \(p(200-5p)=500\) or \(\text-5p^2 + 200p=500\).

Although students are not yet formally introduced to the zero product property, they do have experience with finding the zeros of a quadratic function when given an expression in factored form. This prior knowledge enables them to reason about the solutions to the equation \(p(200-5p)=0\). For instance, noticing the factor \(p\), students are likely to say that one zero of the function is 0.

Ask students,

• “How can you show, without graphing, that \(p=0\) will produce no revenue, or that it is a solution to the equation \(p(200-5p)=0\)?” (When \(p =0\), the factor \(p\) is 0 and therefore the entire expression equals 0.)
• “How can you show that \(p=40\) will also produce no revenue, or that it is also a solution to the same equation?” (When \(p=40\), the factor \((200-5p)\) is 0, and likewise, the entire expression is 0.)
• “Can we use the same reasoning to find the solutions to \(p(200-5p)=500\) or \(\text-5p^2 + 200p =500\)? Why or why not?” (No, not easily. Neither show the zeros of a function. There are many pairs of factors that have 500 as a product.)

Make sure students see that it is fairly straightforward to find the solutions to equations such as \(p(200-5p)=0\), but the same cannot be said about equations such as \(p(200-5p)=500\) or \(\text-5p^2 + 200p =500\).

Highlight that all the equations in this activity are quadratic equations. Explain that a quadratic equation is one that can be written in the form of \(ax^2+bx+c=0\), and where \(a\) is not 0. 

If time permits, ask students to show how all of the equations seen here can be written in this form.