Lesson 9
Solving Quadratic Equations by Using Factored Form
9.1: Why Would You Do That? (10 minutes)
Warmup
In this activity, students find at least one solution of \(x^22x35=0\) by substituting different values of \(x\), evaluating the expression, and checking if it has a value of 0. Experiencing this inefficient method puts students in a better position to appreciate why it may be desirable to write \(x^22x35\) in factored form and use the zero product property.
Launch
Once students have had a chance to evaluate the expression \(x^22x35\) using their chosen number for \(x\), ask if anyone found a value that made the expression equal 0. (It's 7.) Give students a couple of minutes to look for the other value that makes the expression equal 0.
Student Facing
Let's try to find at least one solution to \(x^22x35=0\).
 Choose a whole number between 0 and 10.
 Evaluate the expression \(x^22x35\), using your number for \(x\).
 If your number doesn't give a value of 0, look for someone in your class who may have chosen a number that does make the expression equal 0. Which number is it?
 There is another number that would make the expression \(x^22x35\) equal 0. Can you find it?
Student Response
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Activity Synthesis
If a student found that 5 makes the expression equal 0, ask them to demonstrate that \((\text5)^22(\text5)35\) equals 0.
Discuss with students:
 "Can \(x^22x35\) be written in factored form? What are the factors?" (Yes. \((x7)\) and \((x+5)\))
 "If \(x^22x35\) can be written as \((x7)(x+5)\), can we solve \((x7)(x+5)=0\) instead?" (Yes) "Will the solutions change if we use this equation?" (No. The equations are equivalent, so they have the same solutions.)
 "Why might someone choose to rewrite \(x^22x35=0\) and solve \((x7)(x+5)=0\) instead? (Because the expression is equal to 0, rewriting it in factored form allows us to use the zero product property to find both solutions. It may be more efficient than substituting and evaluating many values for \(x\). It also makes it possible to see how many solutions there are, which is not always easy to tell when the quadratic expression is in standard form.)
9.2: Let’s Solve Some Equations! (15 minutes)
Activity
In this activity, students solve a variety of quadratic equations by integrating what they learned about rewriting quadratic expressions in factored form and their understanding of the zero product property. They begin by analyzing and explaining the steps in a solution strategy, and then applying their observations to solve other equations, both of which require sense making and perseverance (MP1). Students practice attending to precision (MP6) as they study solution steps and communicate what each step does or means.
As students work, notice the equations many students find challenging and those on which errors are commonly made. Discuss these challenges and errors during the activity synthesis.
Launch
Arrange students in groups of 2. Display the worked example in the activity statement for all to see.
Give students 1 minute of quiet time to study the example. Ask them to be prepared to share at least one thing they notice and one thing they wonder. Give students another minute to discuss their observations and questions with a partner.
Students may notice that:
 The right side is equal to 0 after the first step.
 There are two equations starting in step 3.
 The product of 11 and 9 is the constant term in step 2.
Students may wonder:
 why there are two equations in the last two steps
 where the \(\text2n\) went when the expression on the left side was written in factored form
 whether Tyler's work is correct
Then, give students a moment to work quietly on the first question. Encourage students to use relevant mathematical vocabulary in their explanations (such as constant term, squared term, factored form, and zero product property).
Before students proceed to the second set of questions, pause for a class discussion. Ask students to explain each step of Tyler’s solving process and record their explanations for all to see. Encourage students to use reasoning language as opposed to position language: for instance, say "he subtracted 99 from both sides" rather than "he moved the 99 over to the other side."
Make sure students recognize that going from the second line to the third line in Tyler’s work involves finding two factors of 99 that add up to 2. Also discuss why there are two equations at the end. Students should recall that if two numbers multiply to equal 0, then one of the factors must be 0.
Ask students to solve as many equations in the second question as time permits.
Supports accessibility for: Visualspatial processing; Organization
Student Facing

To solve the equation \(n^2  2n=99\), Tyler wrote out the following steps. Analyze Tyler’s work. Write down what Tyler did in each step.
\(\begin {align} n^22n&= 99 &\qquad&\text{Original equation}\\\\n^22n99&=0 &\qquad &\text{Step 1}\\\\ (n11)(n+9)&=0 &\qquad&\text{Step 2} \\\\ n11=0 \quad \text{or} \quad &n+9=0 &\qquad& \text{Step 3}\\\\ n=11 \quad \text{or} \quad &n=\text9 &\qquad&\text{Step 4} \end {align} \)

Solve each equation by rewriting it in factored form and using the zero product property. Show your reasoning.

\(x^2+8x+15=0\)

\(x^28x+12=5\)

\(x^210x11=0\)

\(49x^2=0\)

\((x+4)(x+5)30=0\)

Student Response
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Student Facing
Are you ready for more?
Solve this equation and explain or show your reasoning.
\((x^2x20)(x^2+2x3)=(x^2+2x8)(x^28x+15)\)
Student Response
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Activity Synthesis
Consider displaying the solutions for all to see and discussing only the equations that students found challenging and any common errors.
The last equation is unlike most equations students have seen. Invite students to share how they solved that equation. Discuss questions such as:
 “Can we use the zero product property to write \(x+4=0\) and \(x+5=0\)? Why or why not?” (No. The zero product property can be applied only to products that equal 0. The expression \((x+4)(x+5)30\) has a product for one of its terms, but the expression itself is a difference.)
 “How can it be solved, other than by graphing?” (We can expand the factored expressions using the distributive property and write an equivalent equation: \(x^2 + 9x + 20 30 =0\) or \(x^2 + 9x 10 =0\). This last equation can then be written as \((x+10)(x1)=0\), which allows it to be solved.)
Design Principle(s): Support sensemaking
9.3: Revisiting Quadratic Equations with Only One Solution (10 minutes)
Activity
This activity reinforces what students learned earlier about the connections between the solutions of a quadratic equation and the zeros of a quadratic function. Previously, students were given equations and asked to graph them to determine the number of solutions and their values. Here, they are prompted to work the other way around: to write an equation to represent a quadratic function with only one solution. To do so, students need to make use of the structure of the factored form and their knowledge of the zero product property (MP7).
Launch
Give students access to graphing technology.
Supports accessibility for: Organization; Memory; Attention
Student Facing
 The other day, we saw that a quadratic equation can have 0, 1, or 2 solutions. Sketch graphs that represent three quadratic functions: one that has no zeros, one with 1 zero, and one with 2 zeros.
 Use graphing technology to graph the function defined by \(f(x) = x^2  2x + 1\). What do you notice about the \(x\)intercepts of the graph? What do the \(x\)intercepts reveal about the function?
 Solve \(x^2  2x +1 = 0\) by using the factored form and zero product property. Show your reasoning. What solutions do you get?
 Write an equation to represent another quadratic function that you think will only have one zero. Graph it to check your prediction.
Student Response
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Activity Synthesis
Invite students to share how they solved the equation algebraically. Next, invite students to share the equations they generated. Record and display them for all to see.
Students most likely have written equations in the form of \(g(x)=(x+m)(x+m)\). Ask students why the factored form, rather than the standard form, might have been preferred. Highlight that by using the same expression for the two factors, we know that the solution to \((x+m)(x+m)=0\) will be a single number.
Ask students to describe the graph of a quadratic function with one solution. Point out that this means that the function will have only one zero, and the graph of the function will have a single horizontal intercept.
Design Principle(s): Optimize output; Cultivate conversation
Lesson Synthesis
Lesson Synthesis
Display a series of equations that, prior to this lesson, students could only solve by graphing. For instance:
\(x^2 + 3x 18=0\)
\(x^2  4 = 5x\)
\(x(x7)=\text6\)
\(2x^29x+10=0\)
\((x+6)(x6)=11\)
Ask students to choose an equation that they think they could solve without graphing. Then, ask them to explain to a partner why they believe they could solve the equation.
Consider displaying \((x+9)(x9) = 19\) for all to see and using it as an example: “I think I can solve \((x+9)(x9) = 19\) because I know the expression on the left is equivalent to \(x^29^2\) and I can rewrite the equation as \(x^281 = 19\), which I can solve by rearranging the terms.”
Then, ask if any of the equations appear to be unsolvable other than by graphing and why. Of the equations shown here, \(2x^29x+10=0\) is the only one that students aren’t yet equipped to solve because the coefficient of \(x^2\) is not 1. Students will begin looking at such equations in an upcoming lesson.
9.4: Cooldown  Conquering More Equations (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
Recently, we learned strategies for transforming expressions from standard form to factored form. In earlier lessons, we have also seen that when a quadratic expression is in factored form, it is pretty easy to find values of the variable that make the expression equal zero. Suppose we are solving the equation \(x(x+4)=0\), which says that the product of \(x\) and \(x+4\) is 0. By the zero product property, we know this means that either \(x=0\) or \(x+4=0\), which then tells us that 0 and 4 are solutions.
Together, these two skills—writing quadratic expressions in factored form and using the zero product property when a factored expression equals 0—allow us to solve quadratic equations given in other forms. Here is an example:
\(\displaystyle \begin {align} n^24n &= 140 &\qquad& \text{Original equation}\\n^24n140 & =0 &\qquad& \text{Subtract 140 from each side so the right side is 0}\\ (n14)(n+10) &= 0 &\qquad& \text{Rewrite in factored form} \\ \\n14=0 \quad \text{or} \quad &n+10=0 &\qquad& \text {Apply the zero product property} \\ n=14 \quad \text{or} \quad &n=\text10 &\qquad& \text{Solve each equation}\end{align}\)
When a quadratic equation is written as \(\text {expression in factored form} = 0\), we can also see the number of solutions the equation has.
In the example earlier, it was not obvious how many solutions there would be when the equation was \(n^24n140=0\). When the equation was rewritten as \((n14)(n+10) =0\), we could see that there were two numbers that could make the expression equal 0: 14 and 10.
How many solutions does the equation \(x^220x+100=0\) have?
Let’s rewrite it in factored form: \((x10)(x10)=0\). The two factors are identical, which means that there is only one value of \(x\) that makes the expression \((x10)(x10)\) equal 0. The equation has only one solution: 10.