few_shots: 
Example:
Problem/activity:
Find the general solution for:
$${\mathrm{cos}\left({\pi\over2}+x\right)=\mathrm{tan}{\pi\over4}}$$


Concept/skill: 
Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles.

Answer: 
No. This problem does not directly teach how side ratios in right triangles lead to definitions of trigonometric ratios. Instead, it involves solving a trigonometric function.

Now, does the problem or activity shown below enable students to learn the full intent of the following concept or skill?

shot_info: fl_problem_001894_neg_0

HS_0
standard_description: Generate two numerical patterns using two given rules. Identify apparent relationships between corresponding terms. Form ordered pairs consisting of corresponding terms from the two patterns, and graph the ordered pairs on a coordinate plane. For example, given the rule “Add 3” and the starting number 0, and given the rule “Add 6” and the starting number 0, generate terms in the resulting sequences, and observe that the terms in one sequence are twice the corresponding terms in the other sequence. Explain informally why this is so.
problem_activity: Activity
In this partner activity, students take turns solving equations involving cube roots by isolating the cube root expression and then cubing each side of the equation. As students trade roles explaining their thinking and listening, they have opportunities to explain their reasoning and critique the reasoning of others (MP3).
With square roots, students would have to check that their solutions are valid by substituting in the original equation. That isn’t necessary with cube roots because there is only ever one cube root, but it is a good habit to have.
Launch
Arrange students in groups of 2. After students have had time to identify their three most and least difficult equations, ask students to take turns: the first partner identifies an equation from one of their lists and explains why they chose it, while the other listens and works to understand. Then they switch roles. When both partners have explained why they chose one equation from their most difficult list and one from their least difficult list, they move on to the second question.
When both partners have solved the four equations they chose, ask them to take turns again: the first partner explains how they solved one of their problems, while the other listens and works to understand. When both partners agree on the solution, they switch roles. They should continue to switch roles as time allows.
Conversing: MLR2 Collect and Display.
As students work on this activity, listen for and collect the language students use to explain how they solved the equations. Write the students’ words and phrases on a visual display. As students review the visual display, ask them to clarify the meaning of a phrase. For example, a phrase such as, “Add 4 and then raise to the third power,” can be clarified by restating it as, “Add 4 to each side of the equation. Then raise each side of the equation to the third power.” This will help students use the mathematical language necessary to precisely explain how to solve equations involving cube roots.
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
Engagement: Develop Effort and Persistence.
Encourage and support opportunities for peer collaboration. When students share the equations they selected with a partner, display sentence frames to support conversation such as: “I chose _____ (equation) because . . .”, “I noticed _____ so I chose . . .”, and “I agree/disagree because . . . .”
Supports accessibility for: Language; Social-emotional skills
Student Facing
Here are a lot of equations:
\(\sqrt[3]{t+4} = 3\)
\(\text-10 = \text-\sqrt[3]{a}\)
\(\sqrt[3]{3-w} - 4 = 0\)
\(\sqrt[3]{z} + 9 = 0\)
\(\sqrt[3]{r^3 - 19} = 2\)
\(5 - \sqrt[3]{k+1} = \text-1\)
\(\sqrt[3]{p+4} - 2 = 1\)
\(6-\sqrt[3]{b} = 0\)
\(\sqrt[3]{2n} + 3 = \text-5\)
\(4+\sqrt[3]{\text-m} + 4 = 6\)
\(\text-\sqrt[3]{c} = 5\)
\(\sqrt[3]{s-7} + 3 = 0\)
Without solving, identify 3 equations that you think would be the least difficult to solve and 3 equations that you think would be the most difficult to solve. Be prepared to explain your reasoning.
Choose 4 equations and solve them. At least one should be from your “least difficult” list and at least one should be from your “most difficult” list.
Student Response
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for free access to Student Response.
Student Facing
Are you ready for more?
All of these equations were equivalent to equations that could be written in the form
\(\sqrt[3]{ax+b}+c=0\)
for some constants
\(a\)
,
\(b\)
, and
\(c\)
. Find a formula that would solve any such equation for
\(x\)
in terms of
\(a\)
,
\(b\)
, and
\(c\)
.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
If students have trouble getting started on their more difficult equations, remind them of strategies they used with square roots, such as isolating the root first. Some students may find it helpful to estimate an answer and then check it, in order to get a general idea of what the answer may be and the calculations involved in finding it. If needed, students can look at the graph of
\(y=\sqrt[3]{x}\)
from the previous activity, or
\(y=x^3\)
from earlier in the lesson.
Students who don’t reach a solution should say where they got stuck and what they tried when they share their work with their partner. They can then try finishing the problem together.
Activity Synthesis
Much discussion takes place between partners. Invite students to share how they solved the equations they chose. Here are some questions for discussion:
“How is the process of solving these equations the same as solving equations that involve square roots? How is it different?” (Solving these equations is different because they require cubing rather than squaring. It is also different because cube roots aren’t always positive, so the extra step to check that a positive number isn’t set equal to a negative number isn’t necessary. Solving is the same because it requires isolating the radical expression and then using exponents to make a new equation without radicals.)
“Describe any difficulties you experienced and how you resolved them.” (I cubed a negative number and forgot that the result would be negative; when I plugged in my answer to check it, I saw that I must have done something wrong, so I went back and fixed it. I wasn’t sure how to get started on one of my difficult equations, but I remembered that I should try to isolate the cube root, and then I could see what to do.)


aligns: no