standard_description: Know and apply the Binomial Theorem for the expansion of (x + y)^(n )in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. (The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument.)
problem_activity: Activity
In previous activities and lessons, students found the exact area of a square in order to find an approximation for the square root of an integer. In this activity, students start with a square root of an integer, and draw a square to verify that a given approximation of the square root is reasonable. This is the first time students have seen or drawn squares that do not have vertices at the intersection of grid lines, so it may take them a few minutes to make sense of the new orientation.
Launch
Give students access to four-function calculators. Display the diagram for all to see. Ask students what is the same and what is different about this diagram and diagrams they have seen in earlier activities. Arrange students in groups of 2. Give students 2–3 minutes of quiet work time followed by partner and whole-class discussions.
Action and Expression: Internalize Executive Functions.
Provide students with a graphic organizer for data collection and organizing information.
Supports accessibility for: Language; Organization
Writing, Speaking, Listening: MLR1 Stronger and Clearer Each Time.
After students have had time to identify a point on the number line that is closer to
\(\sqrt{3}\)
, ask them to write a brief explanation of their reasoning. Give students time to meet with 2–3 partners, to share and get feedback on their writing. Display prompts that students can ask each other that will help students strengthen their ideas and clarify their language. For example, “How do you know that
\(\sqrt{3}\)
is between 1.5 and 2?” and “How did you determine the area of the square that you drew?” Students can borrow ideas and language from each partner to refine and clarify their original explanation. This will help students refine both their ideas and their verbal and written output.
Design Principles(s): Optimize output (for explanation); Maximize meta-awareness
Student Facing
Diego said that he thinks that
\(\sqrt{3}\approx 2.5\)
.
[IMG]
Use the square to explain why 2.5 is not a very good approximation for
\(\sqrt{3}\)
. Find a point on the number line that is closer to
\(\sqrt{3}\)
. Draw a new square on the axes and use it to explain how you know the point you plotted is a good approximation for
\(\sqrt{3}\)
.
Use the fact that
\(\sqrt{3}\)
is a solution to the equation
\(x^2 = 3\)
to find a decimal approximation of
\(\sqrt{3}\)
whose square is between 2.9 and 3.1.
Student Response
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Student Facing
Are you ready for more?
A farmer has a grassy patch of land enclosed by a fence in the shape of a square with a side length of 4 meters. To make it a suitable home for some animals, the farmer would like to carve out a smaller square to be filled with water, as in the figure.
What should the side length of the smaller square be so that half of the area is grass and half is water?
[IMG]
Student Response
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Activity Synthesis
Invite one or two students to share their squares. Then tell students, “The square of a point on the number line can be visualized as the area of a literal square. This can help us estimate the value of a square root. Simply squaring the number can as well. Let’s check the squares of some numbers that are potential approximations of
\(\sqrt{3}\)
.”
Then ask students to suggest decimal approximations, and check together as a class by finding their squares. Students should be using each guess to make a better guess next. For example, if they try 1.5, then the square is 2.25, which is too low. This suggests trying bigger. Because we know that 2 is too big (because
\(2^2 = 4\)
, it should be somewhere in between 1.5 and 2.) For example, students might suggest this order:
\(1^2 = 1\)
and
\(2^2=4\)
\(1.5^2 = 2.25\)
\(1.8^2 = 3.24\)
\(1.7^2 = 2.89\)
\(1.72^2 = 2.9584\)
\(1.73^2 = 2.9929\)
So 1.73 is a pretty good approximation of
\(\sqrt{3}\)
.


aligns: no