# Lesson 18

The Quadratic Formula and Complex Solutions

## 18.1: Math Talk: Real or Not? (5 minutes)

### Warm-up

In the previous lesson, students completed the square to decide whether quadratic equations had real or non-real solutions. In this warm-up, students continue to make use of the structure of quadratic expressions to decide whether the solutions are real or non-real (MP7).

### 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 a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

*Representation: Internalize Comprehension.*To support working memory, provide students with sticky notes or mini whiteboards.

*Supports accessibility for: Memory; Organization*

### Student Facing

Mentally decide whether the solutions to each equation are real numbers or not.

\(w^2 = \text- 367\)

\(x^2 +25 = 0\)

\((y + 5)^2 = 0\)

\((z + 5)^2 = \text- 367\)

### Student Response

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### Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. 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?”

*Speaking: MLR8 Discussion Supports.*Display sentence frames to support students when they explain their strategy. For example, "First, I _____ because…” or "I noticed _____ so I….” Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.

*Design Principle(s): Optimize output (for explanation)*

## 18.2: Be Discriminating (15 minutes)

### Activity

Over several lessons, students have developed a concept of complex numbers. In this activity, students use that understanding to consider a situation where the quadratic formula leads to a negative number inside the \(\sqrt{}\) symbol. The focus of the activity is on how complex numbers come into play rather than performing the computation, which is the focus of the next activity.

### Launch

Arrange students in groups of 2. Give a few minutes of quiet think time before asking students to share their ideas with their partner. If there is disagreement, encourage students to work to reach agreement. Follow with whole-class discussion.

*Representing, Conversing: MLR7 Compare and Connect.*Use this routine to prepare students for the whole-class discussion about the solutions to the equation \(3x^2-10x+50=0\). After students solve the equation and plot the solutions in the complex plane, invite them to create a visual display of their work. Then ask students to quietly circulate and observe at least two other visual displays in the room. Give students quiet think time to consider what is the same and what is different about their solution strategies and the placement of the solutions in the complex plane. Next, ask students to find a partner to discuss what they noticed. Listen for and amplify the language students use to explain how to solve the equation and plot the solutions in the complex plane. This will help students explain their thinking as they encounter a situation involving the quadratic formula and complex numbers.

*Design Principle(s): Cultivate conversation*

*Engagement: Develop Effort and Persistence.*Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as: “How did you get. . .?, “I noticed _____ so I . . .”, “Why did you . . .?”, “I agree/disagree because . . . .”

*Supports accessibility for: Language; Social-emotional skills*

### Student Facing

Kiran was using the quadratic formula to solve the equation \(x^2 - 12x + 41 = 0\). He wrote this:

\(\displaystyle x = \dfrac{12 \pm \sqrt{144 - 164}}{2}\)

Then he noticed that the number inside the square root is negative and said, “This equation doesn’t have any solutions.”

- Do you agree with Kiran? Explain your reasoning.
- Write \(\sqrt{\text- 20}\) as an imaginary number.
- Solve the equation \(3x^2 - 10x +50=0\) and plot the solutions in the complex plane.

### Student Response

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### Student Facing

#### Are you ready for more?

Although imaginary numbers let us describe complex solutions to quadratic equations, they were actually discovered and accepted because they could help us find real solutions to equations with polynomials of degree 3. In the 16th century, mathematicians discovered a cubic formula for solving equations of degree 3, but to use it they sometimes had to work with complex numbers. Let’s see an example where this comes up.

- To find a solution to the equation \(x^3-px-q=0\) the cubic formula would first tell us to find a complex number, \(z\), which is \(\frac{q}{2}+i\sqrt{(\frac{p}{3})^3-(\frac{q}{2})^2}\). Find \(z\) when our equation is \(x^3-15x-4=0\).
- The next step is to find a complex number \(w\) so that \(w^3=z\). Show that \(w=2+i\) works for the \(z\) we found in step 1.
- If we write \(w=a+bi\) where \(a\) and \(b\) are real numbers, the solutions to our equation are \(2a\), \(\text-a+b\sqrt{3}\), and \(\text-a-b\sqrt{3}\). What are the three solutions to our equation \(x^3-15x-4=0\)?

### Student Response

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### Activity Synthesis

Select students to share their responses and explain their reasoning. The main takeaway for discussion is that whenever the quadratic formula leads to a negative number inside the \(\sqrt{}\) symbol, it means the original equation has two non-real solutions.

## 18.3: Solving All Kinds of Quadratics (15 minutes)

### Activity

In this row game activity, students solve equivalent quadratic equations given in different forms and then compare their results. The structure of a row game gives students an opportunity to construct viable arguments and critique the reasoning of others (MP3).

Look for groups where one partner uses quadratic formula while the other partner uses a different method for a specific row to share during the discussion. Also look for groups where partners disagree because their solutions look superficially different, for example if one partner gets \(x=2\pm \frac{\sqrt{32}}{2}\) while the other partner gets \(x=2 \pm \sqrt{8}\) for the first question.

### Launch

Arrange students in groups of 2, assigning one student as partner A and the other as partner B. Explain to students that there will be two columns of problems and that they only do the problems in their column. Encourage students to use a different method than their partner on each question and verify that the solutions are equivalent.

*Conversing: MLR8 Discussion Supports.*After students have finished solving their quadratic equation, use this routine to support partner discussion. Invite Partner A to begin with this sentence frame: “To solve, first I _____ because _____. Next I...” or “I tried the method of _____ and what happened was….” Invite the listener, Partner B, to give structured feedback by saying, “I agree/disagree because…”, “Can you explain how you ...?”, or “Another strategy would be _____ because….” This will help students discuss the reasoning involved in solving quadratic equations in different forms.

*Design Principle(s): Support sense-making; Cultivate conversation*

*Engagement: Provide Access by Recruiting Interest.*Leverage choice around perceived challenge. Invite pairs to agree on 3 rows to complete.

*Supports accessibility for: Organization; Attention; Social-emotional skills*

### Student Facing

For each row, you and your partner will each solve a quadratic equation. You should each get the same answer. If you disagree, work to reach agreement.

partner A | partner B |
---|---|

\(x^2 - 4x - 4 = 0 \) | \((x - 2)^2 = 8\) |

\((y - 2)^2 = \text- 8\) | \(y^2 - 4y + 12 = 0 \) |

\((z + \frac32)^2 = \text- \frac{29}{4}\) | \(2z^2 + 6z = \text- 19\) |

\(w^2 + 3w = 5\) | \((w + \frac32)^2 = \frac{29}{4}\) |

\(4t^2-20t+25=0\) | \(4(t^2-5t)= \text-25\) |

### Student Response

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### Anticipated Misconceptions

If students still struggle using imaginary numbers to describe square roots of negative numbers, for example \(\sqrt{\text-8}\), consider asking,

- “What numbers square to make -1?” (\(\pm i\))
- “What numbers square to make 8?” (\(\pm \sqrt{8}\))
- “Then what numbers square to make -8?” (\(\pm i\sqrt{8}\)) Then remind students that, by convention, \(\sqrt{\text-8}\) refers to the number on the positive imaginary axis that squares to make -8, so \(\sqrt{\text-8}=i\sqrt{8}\).

### Activity Synthesis

Select previously identified students to share how they solved their equations and display their work for all to see. Make sure students explain why solutions that look different are actually equal. Ask students “What is the same with the two equations? What is different?” (One of the equations already involves a perfect square, while the other uses the quadratic formula or completing the square to find the solutions. Using the quadratic formula gives the same solution, but will often result in the solutions looking different superficially.)

## Lesson Synthesis

### Lesson Synthesis

Display these equations one at a time and ask students, “Would you solve this with the quadratic formula or by completing the square? Why?”

- \(x^2+6x=4\)
- \(x^2-10x+16=0\)
- \(x^2+5x+8=0\)
- \(\text-2x^2+7x+10 = 0\)
- \(1.3x^2-2.2x=7.8\)

## 18.4: Cool-down - Solve This Quadratic (5 minutes)

### Cool-Down

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## Student Lesson Summary

### Student Facing

Sometimes when we use the quadratic formula to solve a quadratic equation, we get a negative number inside the square root symbol. This means that the solutions to the equation must involve imaginary numbers. For example, consider the following equation:

\(\displaystyle 5x^2 + x + 10 = 0\)

Using the quadratic formula, we know that:

\(\displaystyle x = \dfrac{\text- 1 \pm \sqrt{1^2 - 4 \boldcdot 5 \boldcdot 10}}{2 \boldcdot 5}\)

or

\(\displaystyle x = \dfrac{\text- 1 \pm \sqrt{\text- 199}}{10}\)

Which means that the two solutions are:

\(\displaystyle x = \text- \frac{1}{10} + \frac{\sqrt{199}}{10}i\)

and

\(\displaystyle x = \text- \frac{1}{10} - \frac{\sqrt{199}}{10}i\)