Lesson 17
Completing the Square and Complex Solutions
17.1: Creating Quadratic Equations (5 minutes)
Warm-up
This warm-up prompts students to think about substituting non-real complex numbers for variables in quadratic equations for the first time. Up until this point, students have only substituted real numbers for variables. This opens up the possibilities for what numbers could be solutions to equations. This idea is key for upcoming activities where students solve quadratic equations that have no real solutions but do have non-real solutions.
Student Facing
Match each equation in standard form to its factored form and its solutions.
- \(x^2 - 25 = 0\)
- \(x^2 - 5 = 0\)
- \(x^2 + 25 = 0\)
- \((x - 5i)(x + 5i) = 0\)
- \((x - 5)(x + 5) = 0\)
- \((x - \sqrt{5})(x + \sqrt{5}) = 0\)
- \(\sqrt{5}\), \(\text- \sqrt{5}\)
- 5, -5
- \(5i\), \(\text- 5i\)
Student Response
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Activity Synthesis
Select students to share how they matched equations and solutions. The important idea to discuss is that the kind of numbers that can be solutions to these equations has expanded to include non-real complex numbers.
17.2: Sometimes the Solutions Aren't Real Numbers (10 minutes)
Activity
The purpose of this activity is for students to solve quadratic equations of the sort that result from completing the square, including equations that have solutions that involve imaginary numbers. The left hand side of each equation uses the same expression \((x-5)^2\) so that students can focus on what taking the square root does in each case. In this way, students are primed to make use of structure to see how imaginary numbers arise when solving quadratic equations (MP7).
Launch
Arrange students in groups of 2. Give a few minutes of quiet work time before asking students to share their reasoning with their partner. If there is disagreement, encourage students to work to reach agreement. Follow with whole-class discussion.
Supports accessibility for: Memory; Organization
Student Facing
What are the solutions to these equations?
- \((x - 5)^2 = 0\)
- \((x - 5)^2 = 1\)
- \((x - 5)^2 = \text- 1\)
Student Response
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Activity Synthesis
Select students to explain how they got their solutions.
If not brought up by students, ask students to discuss what is the same and different between the three problems and how these similarities and differences affect the solutions. For \((x-5)^2=0\), there is only one solution because the only number that squares to make 0 is 0. For \((x - 5)^2 = 1\), there are two real solutions because there are two real numbers that square to make 1. For \((x - 5)^2 = \text-1\), there are two non-real solutions because there are two imaginary numbers that square to make -1, and then adding 5 to each side results in non-real, complex solutions.
Design Principle(s): Support sense-making
17.3: Finding Complex Solutions (15 minutes)
Activity
The purpose of this activity is for students to solve two related quadratic equations by completing the square. The process of solving will be nearly identical, with the difference being that one will involve square roots of a negative number and the other will not. It is not necessary for students to prove a general statement about all monic quadratic equations, but it is important for students to compare and understand why some have real solutions and others have non-real solutions.
Launch
Supports accessibility for: Memory; Conceptual processing
Student Facing
Solve these equations by completing the square.
- \(x^2 - 8x + 13 = 0\)
- \(x^2 - 8x + 19 = 0\)
Student Response
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Student Facing
Are you ready for more?
For which values of \(a\) does the equation \(x^2 - 8x + a = 0\) have two real solutions? One real solution? No real solutions? Explain your reasoning.
Student Response
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Activity Synthesis
Select students to share their solutions. Record and display their thinking for all to see. If not brought up during the discussion, ask students to clarify why one equation has real solutions while the other has complex solutions.
Display these three equations for all to see:
- \(x^2+14x+44=0\)
- \(x^2+14x+49=0\)
- \(x^2+14x+54=0\)
Ask students, “How many solutions do these equations have? Which, if any, of these equations has solutions that involve imaginary numbers?” (\(x^2+14x+44\) has two real solutions, \(x^2+14x+49\) has one real solution, and \(x^2+14x+54\) has two non-real solutions. Compare each left hand side to the perfect square \((x+7)^2\), which is equal to \(x^2+14x+49\). When completing the square, the first equation will have \((x+7)^2 = 5\), the second equation will have \((x+7)^2 = 0\), and the third equation will have \((x+7)^2 = \text-5\).)
Design Principle(s): Maximize meta-awareness
17.4: Can You See the Solutions on a Graph? (10 minutes)
Optional activity
This activity is optional because it goes beyond the depth of understanding required by the standards.
The purpose of this activity is to connect graphs of quadratic functions with solutions to related quadratic equations to tell how many solutions an equation has, and whether those solutions are real or non-real. Students have already solved two of the equations in the previous activity, and the equation they haven’t solved involves the perfect square that connects all three of the equations. Graphically, \(y=f(x)\) intersects the \(x\)-axis twice, which corresponds to the two real solutions to \(x^2 - 8x + 13 = 0\). The graph \(y=g(x)\) intersects the \(x\)-axis once, which corresponds to the single real solution to \(x^2 - 8x + 16 = 0\). However, the graph \(y=h(x)\) doesn’t intersect the \(x\)-axis, which reflects the fact that the complex solutions to \(x^2 - 8x + 19 = 0\) can’t be described with a real number line.
Launch
Provide access to graphing technology.
Student Facing
- How many real solutions does each equation have? How many non-real solutions?
- \(x^2 - 8x + 13 = 0\)
- \(x^2 - 8x + 16 = 0\)
- \(x^2 - 8x + 19 = 0\)
- How do the graphs of these functions help us answer the previous question?
- \(f(x) = x^2 - 8x + 13\)
- \(g(x) = x^2 - 8x + 16\)
- \(h(x) = x^2 - 8x + 19\)
Student Response
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Anticipated Misconceptions
Some students may attempt to plot the complex solutions in the coordinate plane as if it were the complex plane. Remind these students that points in the coordinate plane are pairs of real numbers \((x,y)\), whereas points in the complex plane are single complex numbers. In other words, the point \((2,4)\) in the coordinate plane is a different kind of object than the point \(2+4i\) in the complex plane.
Activity Synthesis
Select students to share their solutions and discuss the connections between the graphs and the equations. Discuss the idea that if a graph of a quadratic function doesn’t cross a given horizontal line (in this case, \(x=0\)) this doesn’t indicate that the related equation has no solutions, but rather that it has no real solutions that can be graphed with real number lines.
Design Principle(s): Support sense-making
Lesson Synthesis
Lesson Synthesis
In this lesson, students completed the square to understand when quadratic equations have real or non-real solutions. Here are some questions to consider:
- “How do you know when imaginary numbers will show up when you are solving a quadratic equation?” (Imaginary numbers show up when the process of solving requires finding a number that squares to make a negative number. For example, with \((x-2)^2 = \text- 3\), the quantity \(x-2\) is a number that squares to make -3, so \(x-2 = \pm i\sqrt{3}\).)
- “Without completely solving, how many solutions does \((x-3)^2 - 4 = 0\) have? Are they real solutions or do they involve imaginary numbers?” (It has two real solutions. One way to see this is to recognize that it is in vertex form, and the graph of the function \(f(x)=(x-3)^2 - 4\) is upward facing with a vertex below the \(x\)-axis. That means the graph will intersect the \(x\)-axis in two places, which corresponds to two real solutions.)
- “What about \((x-3)^2+5 = 0\)?” (It has two non-real solutions. One way to see this is to subtract 5 from each side and notice that a quantity squares to make a negative number. Every negative number has two imaginary square roots, so there will be two non-real solutions.)
- “What about \((x-3)^2=0\)?” (There will be one real solution, which is 3. The structure of the expression \((x-3)^2\) reveals that substituting 3 for x will produce a value of 0, and squaring any value other than 0 doesn’t produce 0.)
- “Is it possible for a quadratic equation to have only 1 non-real solution?” (Yes. For example, the equation \((x-i)^2=0\) has exactly one solution, which is \(x=i\).)
17.5: Cool-down - Make One (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
Sometimes quadratic equations have real solutions, and sometimes they do not. Here is a quadratic equation with \(x^2\) equal to a negative number (assume \(k\) is positive):
\(\displaystyle x^2 = \text-k\)
This equation will have imaginary solutions \(i\sqrt{k}\) and \(\text-i\sqrt{k}\). By similar reasoning, an equation of the form:
\(\displaystyle (x- h)^2 = \text-k\)
will have non-real solutions if \(k\) is positive. In this case, the solutions are \(h+ i\sqrt{k}\) and \(h- i\sqrt{k}\).
It isn’t always clear just by looking at a quadratic equation whether the solutions will be real or not. For example, look at this quadratic equation:
\(\displaystyle x^2 - 12x + 41 = 0\)
We can always complete the square to find out what the solutions will be:
\(\displaystyle \begin{align} x^2 - 12x + 36 + 5 &= 0 \\ (x - 6)^2 + 5 &= 0 \\ (x - 6)^2 &= \text- 5 \\ x-6 &= \pm i \sqrt{5} \\ x &= 6 \pm i \sqrt{5} \end{align}\)
This equation has non-real, complex solutions \(6 + i \sqrt{5}\) and \(6 - i \sqrt{5}\).