Lesson 6
Completing the Square
6.1: Fill in the Box (5 minutes)
Warm-up
As a step toward completing the square, students practice identifying the constant term needed to build a perfect square trinomial.
Launch
Arrange students in groups of 2. Give students quiet work time and then time to share their work with a partner.
Student Facing
For each expression, what value would need to be in the box in order for the expression to be a perfect square trinomial?
- \(x^2+10x+\boxed{\phantom{3}}\)
- \(x^2-16x+\boxed{\phantom{3}}\)
- \(x^2+40x+\boxed{\phantom{3}}\)
- \(x^2+5x+\boxed{\phantom{3}}\)
Student Response
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Activity Synthesis
Ask students to compare and contrast the last expression with the others. Was the process of calculating the missing value the same? How could the last expression be rewritten as a squared binomial? (It would look like \(\left(x+\frac52 \right)^2\) or \((x+2.5)^2\).)
6.2: Complete the Process (15 minutes)
Activity
In this activity, students are introduced to completing the square for an equation of a circle. They look at a pre-written version of the first few steps, analyzing what was done and why. Then, they finish the process using skills from previous activities and determine the center and radius of the circle.
Students may be familiar with technology such as the Desmos graphing tool (available under Math Tools) that can graph an equation of a circle. If the technology available in the classroom graphs equations in the form “\(y=\)”, then it will not be a helpful tool for this activity.
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Design Principle(s): Optimize output (for explanation); Cultivate conversation
Supports accessibility for: Memory; Conceptual processing
Student Facing
Here is the equation of a circle: \(x^2+y^2-6x-20y+105 = 0\)
Elena wants to find the center and radius of the circle. Here is what she’s done so far.
Step 1: \(x^2-6x+y^2-20y = \text-105\)
Step 2: \(x^2-6x+9+y^2-20y +100= \text-105+9+100\)
Step 3: \(x^2-6x+9+y^2-20y +100= 4\)
- What did Elena do in the first step?
- Why did Elena add 9 and 100 to the left side of the equation in Step 2?
- Why did Elena add 9 and 100 to the right side of the equation in Step 2?
- What should Elena do next?
- What are the center and radius of this circle?
- Draw a graph of the circle.
Student Response
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Anticipated Misconceptions
If students get stuck, suggest that they look at the form for the equation of a circle that’s on their reference chart. Ask them why that form is useful and how it relates to what they’ve done in recent activities.
Activity Synthesis
Tell students that the process of adding a value to both sides of an equation in order to create a perfect square trinomial is called completing the square. They may have seen a similar process in previous courses.
Here are some questions for discussion:
- “The goal was to rewrite the equation in the form \((x-h)^2+(y-k)^2=r^2\). Why did Elena want to do this?” (If the equation is in this form, we can easily read the center and the radius directly from the equation.)
- “What did Elena do in Step 2, and why?” (She wanted to create perfect square trinomials, so she added the values 9 and 100. To produce an equation equivalent to the original, she also needed to add those values to the other side of the equation.)
- “What did Elena do in Step 1, and why?” (Moving the 105 to the right side makes it easier to create the perfect square trinomials—the 105 would be distracting if it stayed on the left side. Grouping the terms with an \(x\) and those with a \(y\) makes it easier to rewrite the trinomials as squared binomials.)
6.3: Your Turn (10 minutes)
Activity
Students complete the square to find the radius and center of a circle with no scaffolding.
Making graphing technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Student Facing
Here is the equation of a circle: \(x^2+y^2-2x+4y-4=0\)
- Find the center and radius of the circle. Explain or show your reasoning.
- Draw a graph of the circle.
Student Response
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Student Facing
Are you ready for more?
Triangulation is a process of using 3 distances from known landmarks to locate an exact position. Find a point that is located 5 units from the point \((3,4)\), 13 units from the point \((11,\text-4)\), and 17 units from the point \((21,16)\). Use the coordinate grid if it’s helpful.
Student Response
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Anticipated Misconceptions
If students are stuck, suggest they look back to Elena’s process.
Activity Synthesis
Display the image of the circle and two forms of its equation for all to see:
\((x-1)^2+(y+2)^2=3^2\)
\(x^2+y^2-2x+4y-4=0\)
Ask students how these three items relate to each other. (They all represent the same thing in different ways.) Ask students how the concept of distance relates to any of the representations. (The equation \((x-1)^2+(y+2)^2=3^2\) says that the distance between a point \((x,y)\) on the circle and the center \((1,\text-2)\) is 3 units. The graph is a picture of the set of all such points. It’s harder to directly relate the concept of distance to the first equation.)
If students wonder why the equation form given in the activity statement is used at all, tell them that circles are one particular example of a set of shapes called conic sections. All conic sections can be defined by equations similar in form to the one in the activity statement, so sometimes we use that format so we can easily compare and contrast several different equations. They’ll look at another conic section in an upcoming activity.
Supports accessibility for: Visual-spatial processing
Lesson Synthesis
Lesson Synthesis
Display these equations for all to see. Ask students what makes each of these different from the ones they’ve already studied. Then, ask them to find the center and radius of each.
- \(x^2+y^2-24y+143=0\)
- \((x+35)^2+(y-50)^2=19\)
In the first equation, there is no term with a linear \(x\). The expression \(x^2\), however, is in and of itself a perfect square. We could rewrite it as \((x-0)^2\). If we complete the square for the \(y\)-variable, the equation can be written \((x-0)^2+(y-12)^2=1^2\). The center of the circle is \((0,12)\) and the radius measures 1 unit.
The second equation already has squared binomials on the left. We can read that the center of the circle is \((\text-35,50)\). The right side of the equation is not a perfect square. However, it still gives us the radius, which is the positive number that squares to make 19. We can write this as \(\sqrt{19}\). This value is approximately 4.4.
6.4: Cool-down - One More (10 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
Here is an equation for a circle: \(x^2+y^2-4x+6y-3=0\). If we want to find the center and radius of the circle, we can rewrite the equation in the form \((x-h)^2+(y-k)^2=r^2\).
Start by rearranging the terms in the equation to make it easier to work with. Group terms that include the same variable and move the -3 to the right side of the equation.
\(x^2-4x+y^2+6y=3\)
We want the left side to include 2 perfect square trinomials—then, those trinomials can be rewritten in factored form to get the equation in the form we need. To create perfect square trinomials, we can add values to the left side. We’ll keep the equation balanced by adding those same values to the other side.
\(x^2-4x+\boxed{\phantom{3}}+y^2+6y+\boxed{\phantom{3}}=3+\boxed{\phantom{3}}+\boxed{\phantom{3}}\)
For the expression \(x^2-4x\), we need to add 4 to get a perfect square trinomial. For the expression \(y^2+6y\), we need to add 9. Add these values to both sides of the equation. Then, combine the numbers on the right side.
\(x^2-4x+4+y^2+6y+9=3+4+9\)
\(x^2-4x+4+y^2+6y+9=16\)
Now rewrite the perfect square trinomials as squared binomials, and write the 16 in the form \(r^2\).
\((x-2)^2+(y+3)^2=4^2\)
The circle has center \((2,\text-3)\) and radius 4 units.