Lesson 7

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.

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.

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.

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.

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.

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 \(x=\text-2\).

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.

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.

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.

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 \(t\)

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 \(y=\text-\sqrt{t}\).

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.

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.