Lesson 5
How Many Solutions?
5.1: Math Talk: Four Equations (10 minutes)
Warmup
This warmup 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).
Launch
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 wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
Decide whether each statement is true or false.
3 is the only solution to \(x^29=0\).
A solution to \(x^2+25=0\) is 5.
\(x(x7)=0\) has two solutions.
5 and 7 are the solutions to \((x5)(x+7)=12\).
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
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.
Design Principle(s): Optimize output (for explanation)
5.2: Solving by Graphing (15 minutes)
Activity
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.
Launch
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.
Supports accessibility for: Organization; Memory; Attention
Student Facing
Han is solving three equations by graphing.
\((x5)(x3)=0\)
\(\displaystyle (x5)(x3)=\text1\)
\(\displaystyle (x5)(x3)=\text4\)

To solve the first equation, \((x5)(x3)=0\), he graphed \(y=(x5)(x3)\) and then looked for the \(x\)intercepts of the graph.
 Explain why the \(x\)intercepts can be used to solve \((x5)(x3)=0\).
 What are the solutions?

To solve the second equation, Han rewrote it as \((x5)(x3)+1=0\). He then graphed \(y=(x5)(x3)+1\).
Use graphing technology to graph \(y=(x5)(x3)+1\). Then, use the graph to solve the equation. Be prepared to explain how you use the graph for solving.  Solve the third equation using Han’s strategy.

Think about the strategy you used and the solutions you found.
 Why might it be helpful to rearrange each equation to equal 0 on one side and then graph the expression on the nonzero side?
 How many solutions does each of the three equations have?
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
The equations \((x3)(x5)=\text1\), \((x3)(x5)=0\), and \((x3)(x5)=3\) all have wholenumber solutions.

Use graphing technology to graph each of the following pairs of equations on the same coordinate plane. Analyze the graphs and explain how each pair helps to solve the related equation.
 \(y=(x3)(x5)\) and \(y=\text1\)
 \(y=(x3)(x5)\) and \(y=0\)
 \(y=(x3)(x5)\) and \(y=3\)
 Use the graphs to help you find a few other equations of the form \((x3)(x5)=z\) that have wholenumber solutions.
 Find a pattern in the values of \(z\) that give wholenumber solutions.
 Without solving, determine if \((x5)(x3)=120\) and \((x5)(x3)=399\) have wholenumber solutions. Explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
If students enter the equation \((x5)(x3)=0\) 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 \(y=(x5)(x3)\) 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.
Activity Synthesis
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 \((x5)(x3)=\text1\) and the rewritten one \((x5)(x3)+1=0\) 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: \((x4)(x6)=15\)?” (\(y=(x4)(x6)15\)) “What about \((x+3)^21=5\)?" (\(y=(x+3)^26\))
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.”)
5.3: Finding All the Solutions (15 minutes)
Optional activity
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).
Launch
Give students continued access to technology.
Supports accessibility for: Organization; Attention; Socialemotional skills
Student Facing
Solve each equation. Be prepared to explain or show your reasoning.
 \(x^2=121\)
 \(x^231=5\)
 \((x4)(x4)=0\)
 \((x+3)(x1)=5\)
 \((x+1)^2=\text4\)
 \((x4)(x1)=990\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
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 \((x+1)^2=\text4\), 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 \((x+1)^2=\text4\) 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.
5.4: Analyzing Errors in Equation Solving (10 minutes)
Activity
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 \(x^210x\) 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.
Launch
Keep students in groups of 2 and ask them to work quietly on both questions before discussing their responses with a partner.
Design Principle(s): Optimize output (for justification); Cultivate conversation
Student Facing

Consider \((x5)(x+1)=7\). Priya reasons that if this is true, then either \(x5=7\) or \(x+1=7\). So, the solutions to the original equation are 12 and 6.
Do you agree? If not, where was the mistake in Priya’s reasoning?

Consider \(x^2  10x = 0\). Diego says to solve we can just divide each side by \(x\) to get \(x  10 = 0\), so the solution is 10. Mai says, “I wrote the expression on the left in factored form, which gives \(x(x10) = 0\), and ended up with two solutions: 0 and 10.”
Do you agree with either strategy? Explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
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 \(y=x^210x\) 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 \(x^210x\) into the factored form, writing it to equal 0, and solving \(x(x10)=0\) 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.
Lesson Synthesis
Lesson Synthesis
To synthesize the work in this lesson and connect it to prior work, discuss questions such as:
 “How would you solve the equation \(x(x250)=0\)?” (The simplest way would be to use the zero product property, which tells us that one of the factors must be 0, so \(x =0\) and \(x=250\).)
 “If you choose to solve by graphing, is it necessary to rearrange and rewrite the equation first? What equation would you graph and how would you use the graph to solve the equation?” (No. We can just graph \(y=x(x250)\) and look for the \(x\)intercepts of the graph.)
 “Can we use the zero product property to solve \(x(x250)= 100\)? Why or why not?” (No, because the expression on the left does not equal 0.)
 “How would you solve \(x(x250)= 100\)?” (We can rewrite the equation as \(x(x250)  100 =0\), graph \(y=x(x250)100\), and see what the \(x\)intercepts are.)
 “How can the graph tell us how many solutions there are?” (The number of \(x\)intercepts reveals the number of solutions.)
Remind students that examining a graph is not a reliable way to get exact solutions to an equation. For example, the \(x\)intercepts of the graph for \(y=x(x250)100\) are \((\text0.399,0)\) and \((250.399,0)\) and those \(x\)coordinates are likely rounded results.
To solve equations exactly, we need to use algebraic means. In upcoming lessons, we’ll learn more strategies for doing so.
5.5: Cooldown  Two, One, or None? (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
Quadratic equations can have two, one, or no solutions.
We can find out how many solutions a quadratic equation has and what the solutions are by rearranging the equation into the form of \(\text{expression}=0\), graphing the function that the expression defines, and determining its zeros. Here are some examples.

\(x^2=5x\)
Let's first subtract \(5x\) from each side and rewrite the equation as \(x^25x=0\). We can think of solving this equation as finding the zeros of a function defined by \(x^25x\).
If the output of this function is \(y\), we can graph \(y=x^25x\) and identify where the graph intersects the \(x\)axis, where the \(y\)coordinate is 0.
From the graph, we can see that the \(x\)intercepts are \((0,0)\) and \((5,0)\), so \(x^25x\) equals 0 when \(x\) is 0 and when \(x\) is 5.
The graph readily shows that there are two solutions to the equation.
Note that the equation \(x^2=5x\) can be solved without graphing, but we need to be careful not to divide both sides by \(x\). Doing so will give us \(x=5\) but will show no trace of the other solution, \(x=0\)!
Even though dividing both sides by the same value is usually acceptable for solving equations, we avoid dividing by the same variable because it may eliminate a solution.

\((x6)(x4)=\text1\)
Let’s rewrite the equation as \((x6)(x4)+1=0\), and consider it to represent a function defined by \((x6)(x4)+1\) and whose output, \(y\), is 0.
Let's graph \(y=(x6)(x4)+1\) and identify the \(x\)intercepts.
The graph shows one \(x\)intercept at \((5,0)\). This tells us that the function defined by\((x6)(x4)+1\) has only one zero.
It also means that the equation \((x6)(x4)+1=0\) is true only when \(x=5\). The value 5 is the only solution to the equation.

\((x3)(x3)=\text4\)
Rearranging the equation gives \((x3)(x3)+4=0\).
Let’s graph \(y=(x3)(x3)+4\) and find the \(x\)intercepts.
The graph does not intersect the \(x\)axis, so there are no \(x\)intercepts.
This means there are no \(x\)values that can make the expression \((x3)(x3)+4\) equal 0, so the function defined by \(y=(x3)(x3)+4\) has no zeros.
The equation \((x3)(x3)=\text4\) has no solutions.
We can see that this is the case even without graphing. \((x3)(x3) = \text4\) is \((x3)^2=\text4\). Because no number can be squared to get a negative value, the equation has no solutions.
Earlier you learned that graphing is not always reliable for showing precise solutions. This is still true here. The \(x\)intercepts of a graph are not always wholenumber values. While they can give us an idea of how many solutions there are and what the values may be (at least approximately), for exact solutions we still need to rely on algebraic ways of solving.