# Lesson 16

## 16.1: Find the Perfect Squares (5 minutes)

### Warm-up

In this warm-up, students must decide if each quadratic expression is a perfect square. This review of perfect squares will be helpful later in the lesson when students complete the square to solve various quadratic equations.

### Student Facing

The expression $$x^2+8x+16$$ is equivalent to $$(x+4)^2$$. Which expressions are equivalent to $$(x+n)^2$$ for some number $$n$$?

1. $$x^2+10x+25$$
2. $$x^2 + 10x + 29$$
3. $$x^2-6x+8$$
4. $$x^2-6x+9$$

### Activity Synthesis

Select a few students to explain which expressions are perfect squares and how they know. The important idea to review is that the binomial $$(x+n)$$ squares to become $$x^2+2nx+n^2$$, so in order for such a quadratic expression to be a perfect square of a binomial, its constant must be the square of half the coefficient of $$x$$.

## 16.2: Different Ways to Solve It (15 minutes)

### Optional activity

This activity is optional because it revisits content from an earlier course.

The purpose of this activity is for students to analyze two valid methods of solving a quadratic equation (completing the square and the quadratic formula) that result in equivalent answers that look different. This gives students the opportunity to review the two methods. Students make viable arguments and critique the reasoning of others as they inspect the logic of each method and decide whether the different-looking answers are equivalent (MP3).

### Launch

Arrange students in groups of 2. Encourage students to explain their reasoning to their partner as they work, and if there is disagreement, they should work to reach agreement. Remind students of the meaning of $$\pm$$ notation. For example, a result of $$x=2 \pm \sqrt{5}$$ after using the quadratic formula means the original equation has two solutions: $$x=2+\sqrt{5}$$ and $$x=2-\sqrt{5}$$.

Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer collaboration. When students share their work with a partner, display sentence frames to support conversation such as: “First, I _____ because . . .”, “I noticed _____ so I . . .”, “Why did you . . . ?”, “I agree/disagree because . . . .”
Supports accessibility for: Language; Social-emotional skills

### Student Facing

Elena and Han solved the equation $$x^2-6x+7=0$$ in different ways.

Elena said, “First I added 2 to each side:

$$\displaystyle x^2 - 6x + 7 + 2 = 2$$

So that tells me:

$$\displaystyle (x - 3)^2 = 2$$

I can find the square roots of both sides:

$$\displaystyle x - 3 = \pm \sqrt{2}$$

Which is the same as:

$$\displaystyle x = 3 \pm \sqrt{2}$$

So the two solutions are $$x=3+\sqrt{2}$$ and $$x=3-\sqrt{2}$$.”

Han said, “I used the quadratic formula:

$$\displaystyle x = \dfrac{\text- b \pm \sqrt{b^2 - 4 \boldcdot a \boldcdot c}}{2 \boldcdot a}$$

Since $$x^2 - 6x + 7 = 0$$, that means $$a = 1$$, $$b = \text- 6$$, and $$c = 7$$. I know:

$$\displaystyle x = \dfrac{6 \pm \sqrt{36 - 4 \boldcdot 1 \boldcdot 7}}{2 \boldcdot 1}$$

or

$$\displaystyle x = \dfrac{6 \pm \sqrt{8}}{2}$$

So:

$$\displaystyle x = 3 \pm \frac{\sqrt{8}}{2}$$

I think the solutions are $$x = 3 + \frac{\sqrt{8}}{2}$$ and $$x = 3 - \frac{\sqrt{8}}{2}$$.”

Do you agree with either of them? Explain your reasoning.

### Student Facing

#### Are you ready for more?

Under what circumstances would solving an equation of the form $$x^2+bx+c=0$$ lead to a solution that doesn’t involve fractions?

### Anticipated Misconceptions

When analyzing Elena’s work, some students may not remember that $$(x-3)^2=2$$ implies $$x-3=\pm \sqrt{2}$$. Remind these students that for any expression $$A$$, if $$(A)^2=2$$ then the value of $$A$$ could be positive or negative.

### Activity Synthesis

The purpose of the discussion is to establish that Han’s solutions are equivalent to Elena’s, Han correctly used the quadratic formula, and Elena correctly completed the square. Begin the discussion with a few questions to check students' understanding of the two methods, such as:

• “Why does Elena add 2 to each side as her first step?” (Elena adds 2 to each side so that the left hand side will be a perfect square. The expression $$x^2-6x+7$$ is not a perfect square, but $$x^2-6x+9$$ is. Half of the coefficient of $$x$$ is -3, and $$(x-3)^2=x^2-6x+9$$.)
• “When Han used the quadratic formula, why does positive 6 show up even though $$b=\text-6$$?” (Han had to substitute -6 for $$b$$ in the quadratic formula, but the formula itself uses $$\text-b$$. So $$\text-b=\text-(\text-6)=6$$.)

Then, ask students whether they agreed with either Han or Elena and to explain their reasoning. Press students on unsubstantiated claims and include others in the discussion by asking whether they agree. If not mentioned by students, point out that Elena’s solutions look different than Han’s. Ask students how it might be possible for their solutions to look different if they both solved the problem correctly.

Speaking, Representing: MLR8 Discussion Supports. Use this routine to support whole-class discussion. After each student shares, provide the class with the following sentence frames to help them respond: ”I agree because…” or ”I disagree because….” If necessary, revoice student ideas to demonstrate mathematical language use by restating a statement as a question in order to clarify, apply appropriate language, and involve more students. For example, a statement such as, “The answers are the same,” can be restated as a question, “Are Elena’s solutions to the equation $$x^2-6x+7=0$$ equivalent to Han’s solutions?” This will help students engage in the discussion to understand that although solutions look different, it is possible that both are correct.
Design Principle(s): Support sense-making

## 16.3: Solve These Ones (10 minutes)

### Optional activity

This activity is optional because it revisits content from an earlier course.

This activity gives students the opportunity to review solving quadratic equations. Look for students who use different strategies to compare during discussion.

### Launch

Arrange students in groups of 2. Encourage students to share their responses with their partner, and if there is disagreement, work to reach agreement. Encourage students to use a different method than their partner on each question and verify that the solutions are equivalent.

Engagement: Provide Access by Recruiting Interest. Leverage choice around perceived challenge. Invite pairs to agree on 3 of the 6 equations to solve. Chunking this task into more manageable parts may also support students who benefit from additional processing time. Invite students to share their approach with their partner as they finish each problem to support their ability to solve the next equation.
Supports accessibility for: Organization; Attention; Social-emotional skills

### Student Facing

Solve each quadratic equation with the method of your choice. Be prepared to compare your approach with a partner‘s.

1. $$x^2 = 100$$
2. $$x^2 = 38$$
3. $$x^2 - 10x + 25 = 0$$
4. $$x^2 + 14x + 40 = 0$$
5. $$x^2 + 14x + 39 = 0$$
6. $$3x^2 - 5x - 11 = 0$$

### Activity Synthesis

Select 2–3 previously identified pairs to share their strategies and reasons for choosing a particular solution method. If two pairs solved the same problem but got answers that look different, ask the class to verify that their solutions are equivalent.

## Lesson Synthesis

### Lesson Synthesis

In this lesson, students reviewed how to complete the square and use the quadratic formula to solve quadratic equations. Here are some questions for discussion:

• “What situations would make you prefer completing the square and what situations would make you prefer using the quadratic formula?” (Completing the square is efficient in most cases where the coefficient of $$x^2$$ is 1, while the quadratic formula is more easily applied to quadratics with a coefficient of $$x^2$$ not equal to 1.)
• “If one person used one method to get solutions $$x = 5 \pm \sqrt{20}$$ and another person used a different method to get $$x= \dfrac{10 \pm 4 \sqrt{5}}{2}$$, how could you determine whether their solutions are actually the same? (For the fraction, dividing gives $$x=5 \pm 2\sqrt{5}$$. The 5 is the same as the first set of solutions, so they will have exactly the same solutions so long as $$\sqrt{20}$$ is equal to $$2\sqrt{5}$$. In fact, they are equal because they are both positive numbers that square to make 20. I can tell by squaring them: $$(\sqrt{20})^2=20$$ and $$(2\sqrt{5})^2 = 4(\sqrt{5})^2 = 4 \boldcdot 5 = 20$$.)”

## Student Lesson Summary

### Student Facing

$$\displaystyle x^2 - 5x = 25$$

It is often most efficient to solve equations like this by completing the square. To complete the square, note that the perfect square $$(x+n)^2$$ is equal to $$x^2+(2n)x +n^2$$. Compare the coefficients of $$x$$ in $$x^2+(2n)x +n^2$$ to our expression $$x^2-5x$$ to see that we want $$2n=\text-5$$, or just $$n=\text-\frac{5}{2}$$. This means the perfect square $$\left(x-\frac{5}{2}\right)^2$$ is equal to $$x^2 -5x + \frac{25}{4}$$, so adding $$\frac{25}{4}$$ to each side of our equation will give us a perfect square.

\displaystyle \begin{align} x^2 - 5x &= 25 \\ x^2- 5x + \frac{25}{4} &= 25 + \frac{25}{4} \\ \left(x-\frac52\right)^2 &= \frac{100}{4} +\frac{25}{4} \\ \left(x-\frac52\right)^2 &= \frac{125}{4} \end{align}

The two numbers that square to make $$\frac{125}{4}$$ are $$\frac{\sqrt{125}}{2}$$ and $$\text-\frac{\sqrt{125}}{2}$$, so:

$$\displaystyle x-\frac52 = \pm \frac{\sqrt{125}}{2}$$

which means the two solutions are:

$$\displaystyle x = \frac52 \pm \frac{\sqrt{125}}{2}$$

Other times, it is most efficient to use the quadratic formula. Look at the quadratic equation:

$$\displaystyle 3x^2-2x = 0.8$$

We could divide each side by 3 and then complete the square like before, but the equation would get even messier and the chance of making a mistake might be higher. With messier equations like this, it is often most efficient to use the quadratic formula:

$$\displaystyle x = {\text-b \pm \sqrt{b^2-4ac} \over 2a}$$

To use this formula, we first need to put the equation in standard form and identify $$a$$, $$b$$, and $$c$$. Rearranging, we get:

$$\displaystyle 3x^2 - 2x -0.8 =0$$

so $$a=3$$, $$b=\text-2$$, and $$c=\text-0.8$$. We have to be careful to pay attention to the negative signs. Using the quadratic formula, we get:

$$\displaystyle x = {\text-(\text-2) \pm \sqrt{(\text-2)^2-4(3)(\text-0.8)} \over 2(3)}$$

$$\displaystyle x = {2 \pm \sqrt{4+(12)(0.8)} \over 6}$$

Evaluating these solutions with a calculator gives decimal approximations -0.281 and 0.948.