Lesson 4

Solving Quadratic Equations with the Zero Product Property

4.1: Math Talk: Solve These Equations (10 minutes)

Warm-up

This Math Talk introduces students to the zero product property and prepares them to use it to solve quadratic equations. It reminds students that if two numbers are multiplied and the result is 0, then one of the numbers has to be 0. Answering the questions mentally prompts students to notice and make use of structure (MP7).

Launch

Display one problem at a time and ask students to respond without writing anything down. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory; Organization

Student Facing

What values of the variables make each equation true?

\(6 + 2a = 0\)

\(7b=0\)

\(7(c-5)=0\)

\(g \boldcdot h=0\)

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

  • “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
  • “Did anyone have the same strategy but would explain it differently?”
  • “Did anyone solve the problem in a different way?”
  • “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
  • “Do you agree or disagree? Why?”

Highlight explanations that state that any number multiplied by 0 is 0. Then, introduce the zero product property, which states that if the product of two numbers is 0, then at least one of the numbers is 0.

Speaking: MLR8 Discussion Supports. Display sentence frames to support students when they explain their strategy. For example, “First, I _____ because . . .” or “I noticed _____ so I . . . .”  Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class. 
Design Principle(s): Optimize output (for explanation)

4.2: Take the Zero Product Property Out for a Spin (15 minutes)

Activity

In this activity, students solve equations of increasing complexity and do so by reasoning. They begin with linear equations, move toward a series of quadratic expressions in factored form, and end with a cubic expression in factored form. The progression prompts students to reason about the parts and structure of the expressions (MP7), rather than to memorize steps for solving without understanding, and to notice regularity through repeated reasoning (MP8).

As students discuss their reasoning with their partner, listen for those who invoke the zero product property to explain how the last four equations could be solved, and those who notice a pattern in how the equations could be solved. (Though the last question involves a cubic equation, solving it involves the same reasoning as solving quadratic expressions.)

Launch

Arrange students in groups of 2. Tell students to work quietly and answer at least half of the questions before discussing their thinking with a partner.

If needed, remind students that some equations have more than one solution. Because we want students to use reasoning and the structure of equations to develop their solutions, discourage use of graphing technology or spreadsheets in this activity.

Action and Expression: Internalize Executive Functions. To support development of organizational skills, check in with students within the first 2–3 minutes of work time, especially when students arrive at problems with two solutions. Look for students who quickly recognize that they will need to solve two equations and are organizing their work to account for the appropriate number of solutions for each problem. To support students in recognizing examples with two solutions, demonstrate annotating the problem by writing two equations separately, or drawing arrows from each factor to indicate two solutions. Encourage students to prepare their explanations by recording how they recognized the number of solutions as they work.
Supports accessibility for: Memory; Organization

Student Facing

For each equation, find its solution or solutions. Be prepared to explain your reasoning.

  1. \(x-3=0\)
  2. \(x+11=0\)
  3. \(2x+11=0\)
  4. \(x(2x+11)=0\)
  5. \((x-3)(x+11)=0\)
  6. \((x-3)(2x+11)=0\)
  7. \(x(x+3)(3x-4)=0\)

Student Response

For access, consult one of our IM Certified Partners.

Student Facing

Are you ready for more?

  1. Use factors of 48 to find as many solutions as you can to the equation \((x-3)(x+5)=48\).
  2. Once you think you have all the solutions, explain why these must be the only solutions.

Student Response

For access, consult one of our IM Certified Partners.

Anticipated Misconceptions

Students may incorrectly think that \(x\) can represent a different value in each factor in an equation. For example, upon finding -11 and 3 as solutions to \((x-3)(x+11)=0\), they think that one solution is for the \(x\) in \((x-3)\) and the other for the \(x\) in \((x-11)\).

Remind students that solving the equation \((x-3)(x+11)=0\) is like finding the zeros of the function defined by \((x-3)(x-11)\). Although there may be two values of \(x\) that lead to 0 for the value of\((x-3)(x-11)\), only one input can be entered into the function at a time. Ask students to substitute the solutions into the equations and check if the expression is equal to 0 each time.

  • When \(x=\text-11\), the value of the expression is \((\text-11-3)(\text-11 +11)\) or \((\text-14)(0)\), which is 0.
  • When \(x=3\), the value of the expression is \((3-3)(3+11)\) or \((0)(14)\), which is 0.

Activity Synthesis

Invite students to share their strategies for solving the non-linear equations. As they explain, record and organize each step of their reasoning process and display for all to see.

For example, the equation \((x-3)(2x+11)=0\) tells us that, if the product of \((x-3)\) and \((2x+11)\) is 0, then either \(x-3\) is equal to 0, or \(2x+11\) is equal to 0. We can then organize the rest of the solving process as:

If \(x-3\) is equal to 0, then \(x\) is 3.

\(\displaystyle \begin {align} x-3&=0\\ x&=3 \end{align}\)

If \(2x+11\) is equal to 0, then \(x=\text-\frac{11}{2}\)

\(\begin {align} 2x+11&=0\\ 2x &=\text-11 \\x &= \text-\frac{11}{2} \end{align}\)

The equation is true when \(x = 3\) and when \(x=\text- \frac{11}{2}\).

Emphasize that because at least one of the factors must be 0 for the product to be 0, we can write each expression that is a factor to equal to 0 and solve each of these equations separately.

Remind students that we can check our solutions by substituting each one back into the equation and see if the equation remains true. Although the two factors, \((x-3)\) and \((2x-11)\), won’t be 0 simultaneously when 3 or \(\text- \frac{11}{2}\) is substituted for \(x\), the expression on the left side of the equation will have a value of 0 because one of the factors is 0.

  • When \(x\) is 3, the expression is \(\left(0-3\right)\left(2(0)+11\right)\) or \((0)(11)\), which is 0.
  • When \(x\) is \(\text-\frac{11}{2}\), the expression is \(\left(\text-\frac{11}{2}-3\right)\left(2(\text-\frac{11}{2} ) +11\right)\) or \(\left(\text-\frac{17}{2}\right)\left(0\right)\), which is 0.

4.3: Revisiting a Projectile (10 minutes)

Activity

This activity enables students to apply the zero product property to solve a contextual problem and reinforces the idea of solving quadratic equations as a way to reason about quadratic functions.

Previously, students have encountered two equivalent quadratic expressions that define the same quadratic function. Here, they work to show that two quadratic expressions—one in standard form and the other in factored form—really do define the same function. There are several ways to do this, but an efficient and definitive way to show equivalence would be to use the distributive property to expand quadratic expressions in factored form.

Next, they consider which of the two forms helps them find the zeros of the function and then use it to find the zeros without graphing. The work here reiterates the connections between finding the zeros of a quadratic function and solving a quadratic equation where a quadratic expression that defines a function has a value of 0.

Launch

Keep students in groups of 2. Prepare access to graphing technology and spreadsheet tool, in case requested.

Display the two equations that define \(h\) for all to see. Tell students that the two equations define the same function. Ask students how they could show that the two equations indeed define the same function.

Give students a moment of quiet time to think of a strategy and test it, and then time to discuss with a partner, if possible. Then, discuss their responses. Some likely strategies:

  • Graph both equations on the same coordinate plane and show that they coincide.
  • Inspect a table of values of both equations and show that the same output results for any input.
  • Use the distributive property to multiply the expression in factored form to show that \((\text-5t-3)(t-6) =\text-5t^2+27t+18\). (Only this reasoning is really a “proof,” but the other methods supply a lot of evidence that they are the same function.)

Once students see some evidence, ask students to proceed to the activity.

Representation: Internalize Comprehension. Represent the same information through different modalities. Display a sketch of a graph of a projectile, and label the \(y\)-intercept, vertex, and positive \(x\)-intercepts. Keep the display visible for the duration of the activity and refer to it when students discuss why the negative solution is not viable.
Supports accessibility for: Conceptual processing; Language

Student Facing

We have seen quadratic functions modeling the height of a projectile as a function of time.

Here are two ways to define the same function that approximates the height of a projectile in meters, \(t\) seconds after launch:

\(\displaystyle h(t)=\text-5t^2+27t+18 \qquad \qquad h(t)=(\text-5t-3)(t-6)\)

  1. Which way of defining the function allows us to use the zero product property to find out when the height of the object is 0 meters?
  2. Without graphing, determine at what time the height of the object is 0 meters. Show your reasoning.

Student Response

For access, consult one of our IM Certified Partners.

Activity Synthesis

Ask students to share their responses and reasoning. Discuss questions such as:

  • “Why is the factored form more helpful for finding the time when the object has a height of 0 meters?” (To find the input values when the output has a value of 0 is to solve the equation \(\text {quadratic expression}=0\). When the expression is in factored form, we can use the zero product property to find the unknown inputs.)
  • “What if we tried to solve the equation in standard form by performing the same operation to each side?” (We would get stuck. For instance, we could add or subtract terms from each side, but then there are no like terms to combine on either side, so we are no closer to isolating the variable.)

If no students related solving equations in factored form to using the factored form to find the horizontal intercepts of a graph of a quadratic function, discuss that connection.

  • “In an earlier unit, we saw that the factored form of a quadratic expression such as \((x-5)(x+9)\) allows us to see the \(x\)-intercepts of its graph, but we didn’t look into why the graph crosses the \(x\)-axis at those points. Can you explain why it does now?” (The \(x\)-intercepts have a \(y\)-value of 0, which means the quadratic function is 0 at those \(x\)-values: \((x-5)(x+9)=0\). If multiplying two numbers gives 0, one of them must be 0. So either \(x-5=0\) or \(x+9 =0\). If \(x-5=0\), then \(x\) is 5. If \(x+9 =0\), then \(x\) is -9.)
Speaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion. After each student shares, provide the class with the following sentence frames to help them respond: “I agree because . . .” or “I disagree because . . . .” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students. For example, a statement such as, “The first one is more helpful” can be restated as a question such as, “Do you agree that the factored form is more helpful for finding when the object has a height of 0 meters?”
Design Principle(s): Support sense-making

Lesson Synthesis

Lesson Synthesis

To help students consolidate the ideas in the lesson, discuss questions such as:

  • “How does the zero product property help us find the solutions to \((x-3)(x+4)=0\)?” (It tells us that either \(x-3=0\) or \(x+4=0\)”, and each of these equations can be solved easily.)
  • “Can you explain why the solutions to \((x-3)(x+4)=8\) are not 3 and -4?” (The zero product property only works when the product of the factors is zero. When the product is any other number, we can’t conclude that each factor is that number.)
  • “The expression \(x^2-x-12\) is equivalent to \((x+3)(x-4)\). Can we apply the zero product property to solve \(x^2-x-12 = 0\)?” (Only if we rewrite the expression on the left in factored form first. We can’t use the zero product property when the expression is not a product of factors.)
  • “Can we solve \(x^2-x-12 = 0\) by performing the same operation to each side of the equation?” (No, doing that doesn’t help us isolate the variable.)

4.4: Cool-down - Solve This Equation! (5 minutes)

Cool-Down

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Student Lesson Summary

Student Facing

The zero product property says that if the product of two numbers is 0, then one of the numbers must be 0. In other words, if \(a\boldcdot b=0,\) then either \(a=0\) or \(b=0\). This property is handy when an equation we want to solve states that the product of two factors is 0.

Suppose we want to solve \(m(m+9)=0\). This equation says that the product of \(m\) and \((m+9)\) is 0. For this to be true, either \(m=0\) or \(m+9=0\), so both 0 and -9 are solutions.

Here is another equation: \((u-2.345)(14u+2)=0\). The equation says the product of \((u-2.345)\) and \((14u+2)\) is 0, so we can use the zero product property to help us find the values of \(u\). For the equation to be true, one of the factors must be 0.

  • For \(u-2.345=0\) to be true, \(u\) would have to be 2.345.
  • For \(14u+2=0\) or \(14u = \text-2\) to be true, \(u\) would have to be \(\text-\frac{2}{14}\) or \(\text-\frac17\).

The solutions are 2.345 and \(\text-\frac17\).

In general, when a quadratic expression in factored form is on one side of an equation and 0 is on the other side, we can use the zero product property to find its solutions.