# Lesson 7

Inequivalent Equations

## 7.1: 2 and -2 (5 minutes)

### Warm-up

The purpose of this warm-up is to elicit the idea that equations can look similar but have different solutions, which will be useful when students investigate extraneous solutions in a later activity. While students may notice and wonder many things about these equations, the differences and similarities between the solutions to the equations are the important discussion points.

### Launch

Display the list of equations for all to see. Ask students to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice and wonder with their partner, followed by a whole-class discussion.

### Student Facing

What do you notice? What do you wonder?

- \(x^2 = 4\)
- \(x^2 = \text-4\)
- \((x-2)(x+2)=0\)
- \(x=\sqrt{4}\)

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Activity Synthesis

Ask students to share the things they noticed and wondered. Record and display their responses for all to see. If possible, record the relevant reasoning on or near the list of equations. After all responses have been recorded without commentary or editing, ask students, “Is there anything on this list that you are wondering about now?” Encourage students to respectfully disagree, ask for clarification, or point out contradicting information.

If the fact that equations with the same solutions are equivalent to each other does not come up during the conversation, ask students to discuss this idea.

## 7.2: Careful When You Take the Square Root (10 minutes)

### Activity

In this activity, students see that taking the square root of each side of an equation doesn’t necessarily preserve solutions like other algebraic moves they have used in the past.

### Launch

Arrange students in groups of 2. Encourage students to discuss their responses with a partner as they work. If they disagree with their partner, they should work to reach agreement.

*Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to help students improve their written responses for Priya’s question. Give students time to meet with 2–3 partners to share and receive feedback on their responses. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, “How do the graphs show that there are two solutions?”, “What is the definition of a square root?”, and “What happened when Tyler took the square root of each side?” Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students justify why the equation \(x^2-1=3\) must have two solutions.

*Design Principle(s): Optimize output (for justification); Cultivate conversation*

### Student Facing

Tyler was solving this equation:

\(\displaystyle x^2 - 1 = 3\)

He said, “I can add 1 to each side of the equation and it doesn’t change the equation. I get \(x^2 = 4\).”

- Priya said, “It does change the equation. It just doesn’t change the solutions!” Then she showed these two graphs.
- How can you see the solution to the equation \(x^2 - 1 = 3\) in Figure A?
- How can you see the solution to the equation \(x^2 = 4\) in Figure B?
- Use the graphs to explain why the equations have the same solutions.

- Tyler said, “Now I can take the square root of each side to get the solution to \(x^2 = 4\). The square root of \(x^2\) is \(x\). The square root of 4 is 2.” He wrote:
\(\begin{align}x^2 &= 4\\ \sqrt{x^2} &= \sqrt{4} \\ x &= 2 \end{align}\)

Priya said, “But the graphs show that there are

*two*solutions!” What went wrong?

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Activity Synthesis

Invite students to share their responses. Discuss the connection between the graphs and the solutions to the equations. The important idea is that the operation of taking the square root results in a positive number, but there are two square roots of every non-zero number, one positive and one negative. This means that when Tyler takes the square root of both sides of the equation, he does not get an equivalent equation. To find all possible solutions, he should remember that \(x^2\) has two roots, \(x\) and \(\text-x\), and that 4 has two square roots, 2 and -2. This gives him the pair of equations \(x=2\) and \(x=\text-2\).

*Representation: Internalize Comprehension.*Use color-coding and annotations to highlight connections between representations in a problem. As students share their reasoning about the solutions to the equations, scribe their thinking on a visible display. Use the same color to highlight the connection between the graphs and the solutions to the equations.

*Supports accessibility for: Visual-spatial processing; Conceptual processing*

## 7.3: Another Way to Solve (10 minutes)

### Activity

In the previous activity, students saw that taking the square root of both sides of an equation will not preserve solutions to the original equation. This activity gives students an opportunity to consider another type of strategy for solving problems that involve squares and square roots in order to find ways of solving that make sense and that don’t eliminate solutions or introduce extraneous ones.

### Student Facing

Han was solving this equation: \(\displaystyle \dfrac{x+3}{2} = 4\)

He said, "I know that half of \(x+3\) is 4. So \(x+3\) must be 8, since half of 8 is 4. This means that \(x\) is 5."

- Use Han's reasoning to solve this equation: \((x+3)^2 = 4\).
- What advice would you give to someone who was going to solve an equation like \((x+3)^2 = 4\)?

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Student Response.

### Anticipated Misconceptions

If students have trouble getting started on the question about using Han’s reasoning to solve a new problem, it may help them to contrast the way Han solved with a more step-by-step approach like:

\(\displaystyle \frac{x+3}{2}=4\\ x+3=2\boldcdot4 \\x=8-3\\ x=5\)

Ask students to consider what Han did differently from this, or what advice he might give to someone who solved his problem this way. Once students understand Han’s strategy better, it may be easier to apply it to a new problem.

### Activity Synthesis

Show students this graph of the solutions to \((x+3)^2 = 4\):

Ask, “How can we see the solutions to \((x+3)^2 = 4\) in this graph?” (The lines cross at \(x=\text-1\) and \(x=\text-5\), so those are the solutions.)

Invite students to share how they solved the equation. Record student strategies and display them for all to see throughout the discussion.

Then display this for all to see:

\(\displaystyle \begin{align*} (x+3)^2 & = 4 \\ \sqrt{(x+3)^2} & = \sqrt{4} \\ x+3 & = 2 \\ x+3-3 & = 2-3 \\ x & = \text-1 \\ \end{align*}\)

Tell students that this is a way of solving the problem that only results in one of the solutions. Ask, “Why doesn’t this method give us both solutions?” (Taking the square root of both sides only gives us the positive square root.) Then invite students to share their advice for solving equations like this one.

*Speaking: MLR8 Discussion Supports.*As students share their strategies for solving \((x+3)^2=4\), press for details by asking how their reasoning is similar to Han’s, and how they know that \(x+3\) could be 2 or -2. Also ask students how they know that the solutions are \(x=\text-1\) or \(x=\text-5\). This will help students explain how to find both solutions of an equation that involves squares.

*Design Principle(s): Support sense-making; Optimize output (for explanation)*

*Representation: Internalize Comprehension.*Use color and annotations to illustrate student thinking. As students share their reasoning on how they solved the equation, scribe their strategies on a visible display.

*Supports accessibility for: Visual-spatial processing; Conceptual processing*

## 7.4: What Happens When You Square Each Side? (10 minutes)

### Activity

The purpose of this activity is to examine the case in which an equation has no solutions, but squaring each side of the equation results in a new equation that does have a solution. A common mistake is to assume that solutions obtained after squaring each side are solutions to the original equation.

Monitor for students reasoning about why squaring each side of \(\sqrt{x-1} = \text{-} 3\) doesn’t give a solution to the equation by recalling that:

- the \(\sqrt{}\) symbol denotes a positive number, so the equation \(\sqrt{9} = \text- 3\) is false
- the graph of the square root function has no negative outputs, so it’s impossible for \(\sqrt{t}\) to be equal to -3 for any value of \(t\)

### Launch

*Writing, Listening, Conversing: MLR1 Stronger and Clearer Each Time.*Use this routine to help students improve their written responses for the second question. Give students time to meet with 2–3 partners to share and receive feedback on their responses. Display feedback prompts that will help students strengthen their ideas and clarify their language. For example, “What is important to consider here?”, “Can you say that another way?” Invite students to go back and revise or refine their written responses based on the feedback from peers. This will help students refine the advice they give to peers when solving equations with exponents.

*Design Principle(s): Cultivate conversation; Maximize meta-awareness*

### Student Facing

Mai was solving this equation: \(\displaystyle \sqrt{x-1} = 3\)

She said, “I can square each side of the equation to get another equation with the same solutions.” Then she wrote:

\(\begin{align}\sqrt{x-1} &= 3\\ (\sqrt{x-1})^2 &= 3^2 \\ x-1 &= 9 \\ x &= 10 \end{align}\)

- Check to see if her solution makes the original equation true.
- Andre said, “I tried your technique to solve \(\displaystyle \sqrt{x-1} = \text{-} 3\)but it didn’t work.” Why doesn’t it work? Explain or show your reasoning.

### Student Response

### Activity Synthesis

Select previously identified students to share their responses for why squaring each side of \(\sqrt{x-1} = \text{-} 3\) doesn’t give a solution to the equation in this order:

- The solution to \(x-1=9\) is not a solution to \(\sqrt{x-1} = \text{-} 3\) because the equation \(\sqrt{9} = \text-3\) is false. The symbol \(\sqrt{t}\) is used to denote the
*positive*number that squares to make \(t\), so by definition, it can’t be equal to a negative number. - The graph of \(y=\sqrt{t}\) has no negative outputs, so there is no value of \(t\) for which \(\sqrt{t}=\text-3\). In particular, there is no value of \(x\) for which \(\sqrt{x-1} = \text-3\).

Connect these two ideas by discussing that the definition of the \(\sqrt{}\) symbol implies that the graph has no negative outputs. To get negative outputs, the graph would have to be of \(y=\text-\sqrt{t}\).

## 7.5: Solve These Equations With Square Roots in Them (10 minutes)

### Optional activity

This activity is optional because it is an opportunity for extra practice that not all classes may need. Students practice solving equations with a variable inside a square root. There will be more opportunities to practice equations like these in upcoming lessons if time is limited.

### Student Facing

Find the solution(s) to each of these equations, or explain why there is no solution.

- \(\sqrt{t + 4} = 3\)
- \(\text{-}10 = \text{-}\sqrt{a}\)
- \(\sqrt{3 - w} - 4 = 0\)
- \(\sqrt{z} + 9 = 0\)

### Student Response

### Student Facing

#### Are you ready for more?

Are there values of \(a\) and \(b\) so that the equation \(\sqrt{t + a} = b\) has more than one solution? Explain your reasoning.

### Student Response

Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response.

### Activity Synthesis

Ask students to share their responses. After each, invite any students who thought of the problem in a different way to share their response. Remind students that squaring each side of an equation sometimes results in a new equation that has solutions that the old equation doesn’t have.

## Lesson Synthesis

### Lesson Synthesis

In this lesson, students have reasoned about solutions to equations involving squares and square roots. Ask students to solve the equation \((2-x)^2=9\) and share their reasoning. The key idea to discuss is that \(2-x\) could have two possible values because there are two numbers that square to make 9 (namely, 3 and -3). Verify that \(x=\text- 1\) and \(x=5\) are solutions by substituting into the original equation. Then ask students to discuss whether the equations \(\sqrt{2-x}=\text- 3\) and \(2-x=9\) have the same solutions. It is important to note that the equation \(\sqrt{2-x}=\text- 3\) has no solutions because the left hand side is positive and the right hand side is negative. That means that squaring each side can lead to values that aren’t solutions to the original equation. Another way to see this is to solve \(2-x=9\) and substitute its solution into \(\sqrt{2-x}=\text-3\) to show that it is not a solution.

## 7.6: Cool-down - Can You Square It? (5 minutes)

### Cool-Down

Teachers with a valid work email address can click here to register or sign in for free access to Cool-Downs.

## Student Lesson Summary

### Student Facing

Every positive number has *two* square roots. You can see this by looking at the graph of \(y=x^2\):

When we have a square root in an equation like \(\sqrt{t} - 6 = 0\), we can isolate the square root and then square each side:

\(\begin{align}\sqrt{t} - 6 &= 0\\ \sqrt{t} &= 6\\ t &= 6^2 \\ t &= 36 \end{align}\)

But sometimes, squaring each side of an equation gives results that aren’t solutions to the original equation. For example:

\(\begin{align}\sqrt{t} + 6 &= 0\\ \sqrt{t} &= \text-6\\ t &= (\text-6)^2 \\ t &= 36 \end{align}\)

Note that 36 is *not* a solution to the original equation, because \(\sqrt{36}+6\) doesn’t equal 0. In fact, \(\sqrt{t}+6=0\) has no solutions, because it’s impossible for the sum of two positive numbers to be zero.

Remember: sometimes the new equation has solutions that the old equation doesn’t have. Always check your solutions in the original equation!