Lesson 5

This warm-up reminds students of two facts, that in order to use the zero product property, the product of the factors must be 0, and that there is no number that can be squared to get a negative number. At this point, students don’t yet know about complex numbers or that squaring a complex number produces a negative number. Consequently, for now we can just say that there is no number (with a silent “that you know about, yet”) that can be squared to get a negative number.

In explaining why they think each statement is true or false, students practice constructing logical arguments (MP3).

Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and an explanation. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Ask students to share their response and explanation for each problem. Record and display their responses for all to see. After each explanation, give the class a chance to agree or disagree. 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?”

Make sure students understand the rationale that makes each statement true or false, as shown in the student response.

By now, students recognize that when a quadratic equation is in the form of \(\text{expression} = 0\) and the expression is in factored form, the equation can be solved using the zero product property. In this activity, they encounter equations in which one side of the equal sign is not 0. To make one side equal 0 requires rearrangement. For example, to solve \(x(x+6)=8\), the equation needs to be rearranged to \(x(x+6)-8=0\). Yet because the expression on the left is no longer in factored form, the zero product property won’t help after all and another strategy is needed.

Students recall that to solve a quadratic equation in the form of \(\text{expression} = 0\) is essentially to find the zeros of a quadratic function defined by that expression, and that the zeros of a function correspond to the horizontal intercepts of its graph. In the case of \(x(x+6)-8=0\), the function whose zeros we want to find is defined by \(x(x+6)-8\). Graphing \(y=x(x+6)-8\) and examining the \(x\)-intercepts of the graph allow us to see the number of solutions and what they are.

Arrange students in groups of 2 and provide access to graphing technology. Give students a moment to think quietly about the first question and then ask them to briefly discuss their response with their partner before continuing with the rest of the activity.

If students enter the equation \((x-5)(x-3)=0\) into their graphing technology, they may see an error message, or they may see vertical lines. The lines will intersect the \(x\)-axis at the solutions, but they are clearly not graphs of a quadratic function. Emphasize that we want to graph the function defined by \(y=(x-5)(x-3)\) and use its \(x\)-intercepts to find the solution to the related equation. All the points on the two vertical lines do represent solutions to the equation (because the points along each vertical line satisfy the equation regardless of the value chosen for \(y\)), but understanding this is beyond the expectations for students in this course.

Invite students to share their responses, graphs, and explanations on how they used the graphs to solve the equations. Discuss questions such as:

• “Are the original equation \((x-5)(x-3)=\text-1\) and the rewritten one \((x-5)(x-3)+1=0\) equivalent?” (Yes, each pair of equations are equivalent. In that example, 1 is added to both sides of the original equation.)
• "Why might it be helpful to rearrange the equation so that one side is an expression and the other side is 0?" (It allows us to find the zeros of the function defined by that expression. The zeros correspond to the \(x\)-intercepts of the graph.)
• “What equation would you graph to solve this equation: \((x-4)(x-6)=15\)?” (\(y=(x-4)(x-6)-15\)) “What about \((x+3)^2-1=5\)?" (\(y=(x+3)^2-6\))

Make sure students understand that some quadratic functions have two zeros, some have one zero, and some have no zeros, so their respective graphs will have two, one, or no horizontal intercepts, respectively.

Likewise, some quadratic equations have two solutions, some have one solution, and some have no real solutions. (Because students won’t know about numbers that aren’t real until a future course, for now it is sufficient to say “no solutions.”)

This optional activity gives students an opportunity to practice solving quadratic equations and deciding on an effective strategy. Some equations can be easily solved by reasoning. Others would require solving by graphing, because students have not yet learned the strategies to solve algebraically. Students who use graphing technology only when needed practice choosing tools strategically (MP5).

Give students continued access to technology.

Select students to share their solutions and strategies. If not mentioned by students’ explanations, highlight that:

• The first three equations, as well as the equation \((x+1)^2=\text-4\), can be solved by reasoning and that graphing is not necessary.

• The last three equations can be solved by graphing. There are two ways to do so, as shown in a previous activity.

  • One way is graph each side of the equation separately: \(y=\text{expression on one side}\) and \(y= \text{number on the other side}\).

  • Another way is to rearrange the equation such that it is in the form of \(\text{expression} = 0\), graph \(y= \text{expression}\), and then find the \(x\)-intercepts.

• The equation \((x+1)^2=\text-4\) states that some number squared is -4. Because no number can be squared to get a negative number, we can reason that there are no solutions. If this equation is solved by graphing \(y=(x+1)^2+4\), the graph would show no \(x\)-intercepts. This also tells us that there are no solutions.

This activity aims to uncover some common misconceptions in solving quadratic equations and to reinforce that certain familiar moves for solving equations are not effective. Students critique several arguments on how to solve quadratic equations. In articulating why certain lines of reasoning are correct or incorrect, they practice constructing logical arguments (MP3).

As students work, look for students who:

• explain both the error in Priya’s argument and the validity of Mai’s argument in terms of the zero product property
• notice that Diego’s method disregards the second solution of the equation
• create and use a graph to verify their critique of Priya, Mai, or Diego's work (for example, graphing to show that \(x^2-10x\) has two \(x\)-intercepts)

Multiplying or dividing both sides of an equation by a variable expression can change the solution set of an equation, either by eliminating a solution (as shown in Diego's method) or introducing a new solution (for example, starting with the equation \(x=3\) and multiplying both sides by \(x\) to get \(x^2=3x\) gives an equation that now has 2 solutions). Students will learn more about such moves in a later course.

Keep students in groups of 2 and ask them to work quietly on both questions before discussing their responses with a partner.

Select previously identified students to share their responses and reasoning. Here are some key observations to highlight:

• For the first question, make sure students understand that the zero product property only works when the product of the factors is 0. While substituting 6 for \(x\) in the expression does produce 7, substituting 12 does not give the same result.
• For the second question, consider graphing the function \(y=x^2-10x\) so students can see that the graph intersects the \(x\)-axis at two points, which means that there are two \(x\)-values that give a zero output: 0 and 10. Rewriting \(x^2-10x\) into the factored form, writing it to equal 0, and solving \(x(x-10)=0\) allow us to see that this is indeed the case.
• Dividing each side by a variable (as what Diego did) seems to enable us to isolate the remaining variable, but only one solution remains. Dividing each side of an equation by \(x\) is not a valid move because when \(x\) is 0, the expressions on each side become undefined.