Lesson 14
Completing the Square (Part 3)
14.1: Perfect Squares in Two Forms (5 minutes)
Warmup
In this warmup, students consider perfect squares in standard form in which the leading coefficient is not 1. They use what they know about expanding \((x+m)^2\) to analyze the expansion of an expression of the form \((kx + m)^2\) and to identify an error. In explaining and correcting the error, students practice constructing logical arguments (MP3).
Launch
Display the entire task for all to see. Give students 2 minutes of quiet think time. Select students to share their responses and how they reasoned about the error in Elena’s statement.
Student Facing
Elena says, “\((x+3)^2\) can be expanded into \(x^2 + 6x + 9\). Likewise, \((2x+3)^2\) can be expanded into \(4x^2 + 6x + 9\).”
Find an error in Elena’s statement and correct the error. Show your reasoning.
Student Response
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Activity Synthesis
Select 1–2 students to share their responses. Make sure students see that expanding the expression \((kx + m)^2\) involves the same structure as expanding \((x+m)^2\). Highlight the structure using a diagram and the distributive property.
\(x\)  \(3\)  

\(x\)  \(x^2\)  \(3x\) 
\(3\)  \(3x\)  \(9\) 
\(2x\)  \(3\)  

\(2x\)  \(4x^2\)  \(6x\) 
\(3\)  \(6x\)  \(9\) 
When we expand \((x+3)^2\), the squared term is \(x^2\), the linear term is 2 times \((3)(x)\), which is \(6x\), and the constant term is \(3^2\). When we expand \((2x+3)^2\), the squared term is \((2x)^2\), the linear term is 2 times \((2x)(3)\), which is \(12x\), and the constant term is \(3^2\).
Students will have opportunities to transform more of such expressions later and, with practice, will better see the effects of squaring a linear expression like \(2x+3\), in which the coefficient of the variable is not 1.
14.2: Perfect in A Different Way (15 minutes)
Activity
In this activity, students use the structure they saw earlier to rewrite squared expressions of the form \((kx+m)^2\), where \(k\) is not 1, into standard form. Through repeated reasoning, students see how the coefficient \(k\) is related to the values of \(a\) and \(b\) when the expression is expanded into \(ax^2+bx+c\), and how the \(m\) is related to \(c\) (MP8).
As students work, monitor for the following strategies for rewriting the expressions in the first question (from factored form to standard form):
 Drawing rectangular diagrams.
 Applying the distributive property, writing out all the expanded terms, and combining like terms.
 Generalizing that when expanding an expression such as \((4x+1)^2\):
 The coefficient 4 in \(4x+1\) gets squared, so \(a\) in \(ax^2 + bx+c\) is \(4^2\).
 The coefficient 4 in \(4x+1\) gets multiplied by the constant term 1 twice, so \(b\) is \(2(4)(1)\).
 The constant term 1 in \(4x+1\) is squared, so \(c\) is \(1^2\).
Identify students using each strategy and invite them to share during the class discussion.
Launch
Explain to students that they will now practice squaring expressions such as \(4x+1\) and \(\frac12 x+7\), where the coefficient of the linear term is not 1, and writing them in standard form.
Consider arranging students in groups of 2 and asking them to think quietly about each question before conferring with their partner.
Student Facing

Write each expression in standard form:
 \((4x+1)^2\)
 \((5x2)^2\)
 \((\frac12 x + 7)^2\)
 \((3x+n)^2\)
 \((kx+m)^2\)

Decide if each expression is a perfect square. If so, write an equivalent expression of the form \((kx+m)^2\). If not, suggest one change to turn it into a perfect square.
 \(4x^2 + 12x + 9\)
 \(4x^2 + 8x+ 25\)
Student Response
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Activity Synthesis
Select previously identified students to share their responses and strategies for the first question, starting with students who used diagrams and ending with students who generalized the pattern in the rewriting process. As students explain, display or record their reasoning for all to see.
If not mentioned in students’ explanations, highlight the generalization noted in the activity narrative.
Then, discuss how students used their insights from the first question to help them identify perfect squares, or to turn expressions into perfect squares, in the second question. Make sure students see the structure behind the values of \(a\), \(b\), and \(c\) when the expression \(ax^2+bx+c\) is a perfect square. This structure will help them complete the square in the next activity.
Design Principle(s): Maximize metaawareness; Cultivate conversation
Supports accessibility for: Language; Organization
14.3: When All the Stars Align (15 minutes)
Activity
Earlier, students noticed that when an expression \(kx+m\) is squared and written in standard form \(ax^2+bx+c\), the value of \(b\) is \(2km\) and the value of \(c\) is \(m^2\). In this activity, they use these observations to complete expressions to make them perfect squares and then write them as squared factors. Then, they apply this skill to solve quadratic equations.
Launch
To reiterate the connections between \((kx + m)^2\) and its equivalent expression of the form \(ax^2+bx+c\), display these two expressions for all to see and ask students:
 “If the two expressions are perfect squares and are equivalent, how is \(a\) related to \(k\)?” (\(a\) is \(k^2\).)
 “How is \(c\) related to \(m\)?” (\(c\) is \(m^2\).)
 “How is \(b\) related to \(k\) and \(m\)?” (\(b\) is \(2km\).)
Tell students they will use these insights to complete some squares. Consider keeping students in groups of 2.
Supports accessibility for: Visualspatial processing; Conceptual processing
Student Facing
 Find the value of \(c\) to make each expression in the left column a perfect square in standard form. Then, write an equivalent expression in the form of squared factors. In the last row, write your own pair of equivalent expressions.
standard form \((ax^2+bx+c)\) squared factors \((kx+m)^2\) \(100x^2+80x+c\) \(36x^260x+c\) \(25x^2+40x+c\) \(0.25x^214x+c\) 
Solve each equation by completing the square:
\(25x^2 + 40x = \text12\)
\(36x^2  60x + 10 = \text6\)
Student Response
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Anticipated Misconceptions
If students struggle to follow the generalized relationships between \((kx + m)^2\) and its equivalent expression of the form \(ax^2+bx+c\) (as discussed in the launch), consider revisiting a concrete example from an earlier activity—for example, \((5x2)^2\). Ask students to relate each term in the squared factor to the terms of its expression in standard form, \(25x^2  20x + 4\).
If further scaffolding is helpful, consider using an example with an additional variable (for instance, \((3x+n)^2\) and its counterpart in standard form, \(9x^2 + 2(3)(n)x + n^2\)) before generalizing the equivalence of the two forms in entirely abstract terms.
Activity Synthesis
Ask students to write the standardform expression they invented on a piece of scrap paper, except without the constant term. (From the blank row of the table.) Invite them to switch papers with a partner, and complete the square and write each other’s expressions in factored form. If they get stuck, encourage them to talk with their partner to work together and fix any mistakes.
If time permits, invite students to share their responses and strategies. Discuss questions such as:
 “How did you know what value to use for \(c\) in, say, \(100x^2 + 80x + c\)?” (\(k^2=100\), so \(k\) is either 10 or 10. If \(k\) is 10, then \(2(10)m=80\) and \(m=4\). We know that \(c=m^2\), so \(c\) is \(4^2\) or 16. If \(k\) is 10,then \(2(\text10) m=80\) and \(m= \text4\). Squaring 4 still gives 16.)
 “How did you know whether \(c\) would be positive or negative?” (\(c\) is some number squared, so it will always be positive.)
 “In the equation \(36x^2  60x + 10 = \text6\), the expression on the left already has 10 as a constant term but it is not a perfect square. How do we make it a perfect square?” (25 would make it a perfect square, so we can either add 15 to each side of the equation, or first subtract 10 from each side and then add 25.)
As students explain their solution methods on the second question, record and display their reasoning for all to see, or display a worked solution such as:
\(\begin{align} 36x^2  60x + 10 &= \text6\\ 36x^2  60x + 10 + 15 &= \text6 + 15\\ 36x^2  60x + 25 &= 9\\ (6x5)^2 &=9 \end{align}\)\( \)
\(\begin {align} 6x5 =3 \quad &\text{or} \quad 6x5 =\text3\\ 6x = 8 \quad &\text{or} \quad 6x = 2\\ x = \frac86 =\frac43 \quad & \text{or} \quad x = \frac13 \end{align}\)
14.4: Putting Stars into Alignment (30 minutes)
Optional activity
This activity is optional. It shows three different methods for solving an equation: by rewriting it in factored form, by transforming it into an expression in which the squared term has a coefficient 1 and then rewriting in factored form, and by completing the square. The second method involves temporarily substituting a part of the expression with another variable so that the squared term has a coefficient of 1, which makes it easier to rewrite in factored form. Students encountered this strategy in an optional activity (“Making It Simpler”) in an earlier lesson. If students did not do that optional activity, consider doing it first before using this activity.
The numbers in the equations here do not lend themselves to be easily rewritten in factored form (without guessing and checking or substitution), or to be easily rewritten into perfect squares (because the leading coefficient and the constant terms are not perfect squares), but students see that these equations can be solved with these strategies. The activity further illustrates that any of these methods can be rather tedious, providing further motivation to solve using the quadratic formula, which students will learn in a future lesson.
Launch
Display the perfect squares students saw in the preceding activity. Ask them what they notice about the coefficient of the square term in each expression.
\(\displaystyle 100x^2 + 80x + 16\)
\(\displaystyle 36x^2  60x + 25\)
\(\displaystyle 25x^2 + 40x + 16\)
\(\displaystyle 0.25x^2 14x + 196\)
Students are likely to notice that the coefficient of every \(x^2\) is a perfect square, which made it easier to rewrite the completed square as squared factors. Solicit some ideas about whether we could still complete the square if a quadratic expression does not have a perfect square for the coefficient of \(x^2\). Then, tell students that they will explore a few methods for solving such equations.
Arrange students in groups of 2. Ask one partner to study the first two methods and the other partner to study the third method, and then take turns explaining their understanding to each other. Discuss the methods, especially the third one, before students begin using them to solve equations.
Make sure students see that, in both the second and third methods, the first step involves multiplying both sides of the equation by 3 to make the coefficient of \(x^2\) a perfect square. Each term on the left side of the equation changes by a factor of 3, but the right side of the equation remains 0 because multiplying 0 by any number results in 0.
If time is limited, consider asking students to solve only three equations, using each method once.
Supports accessibility for: Memory; Organization
Student Facing
Here are three methods for solving \(3x^2 + 8x + 5 = 0\).
Try to make sense of each method.
Method 1:
\(\displaystyle \begin {align}3x^2 + 8x + 5 &= 0\\ (3x + 5)(x + 1) &= 0 \end{align}\)
\(\displaystyle \begin {align} x = \text \frac53 \quad \text{or} \quad x = \text1\end {align}\)
Method 2:
\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\ (3x)^2 + 8(3x) + 15 &= 0\\ U^2 + 8U + 15 &= 0\\ (U+5)(U+3) &= 0 \end{align}\)
\(\displaystyle \begin {align} U = \text5 \quad &\text{or} \quad U = \text3\\3x = \text5 \quad &\text{or} \quad 3x = \text3\\ x = \text \frac53 \quad &\text{or} \quad x = \text1 \end{align}\)
Method 3:
\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\9x^2 + 24x + 16 &= 1\\(3x + 4)^2 &= 1 \end{align}\)
\(\displaystyle \begin {align} 3x+4 = 1 \quad & \text{or} \quad 3x+4 = \text1\\x = \text1 \quad & \text{or} \quad x = \text \frac53 \end {align}\)
Once you understand the methods, use each method at least one time to solve these equations.
 \(5x^2 + 17x + 6 = 0\)
 \(6x^2 + 19x = \text10\)
 \(8x^2  33x + 4 = 0\)
 \(8x^2  26x = \text21\)
 \(10x^2 + 37x = 36\)
 \(12x^2 + 20x  77=0\)
Student Response
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Student Facing
Are you ready for more?
Find the solutions to \(3x^2 6x + \frac{9}{4} = 0\). Explain your reasoning.
Student Response
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Activity Synthesis
Consider asking students who solve the same equation using different methods to compare and contrast their solution strategies. Then, invite them to reflect on the solving process:
 “Did the numbers in certain equations make it easier to solve by one method versus the others?”
 “What are the advantages of each method? What about its drawbacks?”
 “Is there a strategy you prefer or find reliable? Which one and why?”
Tell students that in upcoming lessons, they will look at a more efficient method for solving quadratic equations.
Design Principle(s): Support sensemaking
Lesson Synthesis
Lesson Synthesis
Invite students to reflect on the process of completing the square for various kinds of equations. Display equations such as the following and ask students which ones would be fairly easy to solve by completing the square, which ones would not be, and why.
\(\displaystyle x^2 +8x = \text12\)
\(\displaystyle 16x^2 + 8x = \text12\)
\(\displaystyle 4x^2 + 32x = \text7\)
\(\displaystyle 5x^2 + 32x = \text7\)
\(\displaystyle 9x^2 + 3x  6 = 0\)
Discuss questions such as:
 “What number would you add to \(4x^2 + 32x\) (the third equation on the list) to make it a perfect square? How do you know?” (64. The coefficient 32 is equal to \(2(2)m\), so \(m\) is 8. The number to be added is \(8^2\), which is 64.)
 “What about \(9x^2 + 6x\)? How do you know?” (1. The coefficient 6 is equal to \(2(3)m\), so \(m\) is 1. The number to be added is \(1^2\), which is 1.)
 “Do certain features or numbers in an equation make it easier or harder to solve it by completing the square? If so, which features or what kinds of numbers?” (It’s easier when the coefficient of the squared term is 1 or another perfect square. It’s harder when some of the coefficients are fractions or the quadratic term’s coefficient is not a perfect square.)
14.5: Cooldown  One More Equation (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
In earlier lessons, we worked with perfect squares such as \((x+1)^2\) and \((x5)(x5)\). We learned that their equivalent expressions in standard form follow a predictable pattern:
 In general, \((x+m)^2\) can be written as \(x^2 + 2mx + m^2\).
 If a quadratic expression is of the form \(ax^2 + bx + c\), and the value of \(a\) is 1, then the value of \(b\) is \(2m\), and the value of \(c\) is \(m^2\) for some value of \(m\).
In this lesson, the variable in the factors being squared had a coefficient other than 1, for example \((3x+1)^2\) and \((2x5)(2x5)\). Their equivalent expression in standard form also followed the same pattern we saw earlier.
squared factors  standard form 

\((3x+1)^2\)  \((3x)^2 + 2(3x)(1) + 1^2 \quad \text{or} \quad 9x^2 +6x + 1\) 
\((2x5)^2\)  \((2x)^2 + 2(2x)(\text5) + (\text5)^2 \quad \text{or} \quad 4x^2 20x + 25\) 
In general, \((kx+m)^2\) can be written as:
\(\displaystyle (kx)^2 + 2 (kx)(m) + m^2\)
or
\(\displaystyle k^2 x^2 + 2kmx + m^2\)
If a quadratic expression is of the form \(ax^2 + bx + c\), then:
 the value of \(a\) is \(k^2\)
 the value of \(b\) is \(2km\)
 the value of \(c\) is \(m^2\)
We can use this pattern to help us complete the square and solve equations when the squared term \(x^2\) has a coefficient other than 1—for example: \(16x^2 + 40x = 11\).
What constant term \(c\) can we add to make the expression on the left of the equal sign a perfect square? And how do we write this expression as squared factors?
 16 is \(4^2\), so the squared factors could be \((4x + m)^2\).
 40 is equal to \(2(4m)\), so \(2(4m) = 40\) or \(8m=40\). This means that \(m = 5\).
 If \(c\) is \(m^2\), then \(c = 5^2\) or \(c=25\).
 So the expression \(16x^2 + 40x + 25\) is a perfect square and is equivalent to \((4x+5)^2\).
Let’s solve the equation \(16x^2 + 40x = 11\) by completing the square!
\(\displaystyle \begin {align} 16x^2 + 40x &= 11\\ 16x^2 + 40x + 25 &= 11 + 25\\ (4x + 5)^2 &=36\\\\4x+5 = 6 \quad &\text {or} \quad 4x+5= \text6\\ 4x = 1 \quad &\text {or} \quad 4x = \text11\\ x=\frac14 \quad &\text {or} \quad x = \text \frac{11}{4} \end {align}\).