Thursday, April 28, 2016

Day 148: Pythagorean Theorem Proof #2 & Intersection Posters

My 2nd period class had the best estimates of the day in my opinion. Katie knew that from San Francisco to Florida it's about 2400 miles. She reasoned that from Pennsylvania to Florida it would be half that, so she estimated around 1200 miles. This was a pretty accurate estimate. Estefany reasoned that today's distance was 3 times longer than yesterday's distance of 271. 
First attempt at explaining the what converse meant... not so good.
Gary did an awesome job showing his work when finding the distance between 2 coordinates using the pythagorean theorem and showing all his work.
The second proof we did today was now starting with two squares with a total area of a^2 + b^2. Instead of using a ruler to measure side b, I had students tear off a strip from their half sheet of paper to make marks for side length B. They then slid it to the right and marked a leg of length b, and then connected that with a dashed diagonal line to the top left to form a right triangle with hypotenuse of length c.

Students then reasoned that the rest of the base must be A, because the whole top is a+b and a+b-b is equal to a. Then they drew a diagonal from leg A to leg b to form a second right triangle.

To avoid any mistakes, I got a highlighter and outlined the border, and made sure to point out where I did not want them to cut (the line where the square a and b are adjacent to each other. They then cut this, after being sure to label side c. In this example I forgot to label the second side c.

After this I instructed them to cut the 2 triangles out by staying on the pink line. They then rearranged it to form a square. Some used the 2 triangles to make a rectangle, and I asked them if that was a square. I encouraged them to use all 3 pieces. After some rotations, it fit together. I asked them what the area of the new square was and they said c^2. They realized the hypotenuse was the side length. A lot of students appreciated the elegance of this proof.

My best attempt at explaining the converse. In the top right, we stated the pythagorean theorem followed by its converse.

Day 147: Pythagorean Theorem Proof 1 & More Systems


I don't recall which flight distance we estimated this day.

Students worked on their first proof of the Pythagorean Theorem. You start with 4 congruent right triangles with sides labeled a, b, and c. They arrange them with the 90 degree angles matched up with the corners of a square. They are asked why is the middle, empty, unshaded area c^2? I had to poke and prod students to go further with their explanations. Each triangle had its hypotenuse, c, facing in, to form 4 congruent sides of a square with 4 90 degree angles and 4 sides of length c. To find the area, they said length times width or base times height to get c*c giving them c squared.

The text prompts them to reason if the area within the square will change if you move the triangles around. They reasoned that the area of the triangles hadn't changed and neither had the square, so the area wouldn't change. They then rearranged it like the second photo below this one...

They were asked why the empty or unshaded area was a^2+b^2. I liked when students reasoned that the hypotenuses were facing each other so the sides of b length were exposed to form the smaller square of area b^2. The smaller square had side lengths of a, making it a^2. Therefore they concluded, that the original square, c squared, was equal to the sum of a^2 plus b^2. I asked students if they were like terms, or were the same variable. They said no, which is why we can't add them together as variables.

In accelerated students grappled with a challenging system that was an exponential equation and a linear equation. It couldn't be solved algebraically, though some students were determined and tried figuring it out with a TI calculator. I encouraged students to graph both functions to find the intersections. I saw that some students graphs of 2^x-3 were looking like parabolas, so I stopped the class to review negative exponents by reviewing exponents of base 2.


I wanted to make sure they remembered the relationship between 2 squared and 2 to the -2nd power. They are reciprocals of each other.

Then students worked on sketches of possible intersections. As you can see below, one situation called for 2 parabolas with 1 point of intersection. Many students thought of a parabola and an upside down reflected parabola. Students were impressed that Aaron came up with 2 side by side parabolas that intersected only once.


One student pointed out that a line and a parabola that intersect only once would be called a tangent line.

Then later that evening I tutored an Algebra 2 student on Graphing the Sine function. He challenged himself to do the same function except with the cosine function. This is a great activity by Shelly Carranza. Go to teacher.desmos.com and search "Graphing the Sine Function."

Wednesday, April 27, 2016

Day 146: Fish Tale Pythagorean 3D (Notice, Wonder, Q's)

Today students noticed and wondered about the Fish Tale problem. I used it from Yummymath.com last year, but this time I didn't give the numbers or the wording to the problem, and asked them what they noticed, wondered, and any questions they had. The results were fantastic. It seems like more students were successful this way then last year. Interestingly, more students solved it the simpler way this year of seeing the 3 dimensional right triangle rather than the method that requires 1 more step using the Pythagorean Theorem.

Below I've embedded the Google Slides.  Slides 8, 9, and 10 were the slide duplicated from each period so that I could refer back to it. I wrote what students noticed, what they wondered, and their questions.

Sometimes when they ask questions, I'll immediately answer them by typing into parentheses () right after they ask it. I like how they noticed an unrealistic nuance of the problem: Betty is a fresh water gold fish and Stripe is a salt water clownfish (Nemo). I liked how students were interpreting the lines as the swimming paths and added some more details. After doing Taco cart, students asked if the fish swam at different speeds. I think I could add that element in next year as a sort of extension. I do want them more to focus on how to show their work so that a class mate that didn't understand it could by looking at their work.

An example of the 3 step process is below. I like how Jonathan drew another diagram in the bottom right to show how he saw the 3rd right triangle, whose leg was the diagonal of the floor of the rectangular base.
Jonathan's detailed work.
Afterwards we reviewed a word problem that had to be interpreted with a diagram, and then solved for the missing leg. The cliff was 75 feet tall, and a 96 foot rope was pulled tight. You had to figure out how far the rope was from the base of the cliff and where Mark was standing.

This example by Bing in 4th period was great. He explained his work as he was writing it and got a nice applause.
I scaffolded to students how to properly draw a rectangular prism with a dashed line to show the faces behind it that you couldn't see. I also asked them to color code the paths of each fish.
Madison shared how she saw the 3 dimensional right triangle by peering in from the left hand face and seeing stripes purple path going from the bottom front left to the back top right. It's easy to see because she also 

I didn't take a picture of accelerated but we used Desmos to confirm solving the system of equations and how the graphs reflected the points of intersections. Students worked better in their groups today than yesterday. They figured out that intercepts were where functions crossed the x or y axes and had a 0 as the x or y coordinate. Interceptions were where two functions crossed paths. They realized it wasn't just lines that could cross but a parabola and a linear function.

Tuesday, April 26, 2016

Day 145: Pythagorean Theorem Learning Goals / Success Criteria & Intersections of Quadratic & Linear Functions

Today was the second day of estimating plane flight distances. The benchmark was a cross country flight from Los Angeles to Virginia. That was 2227 miles. Students reasoned the flight to Chicago would be one fourth the distance so they divided the previous distance by 4. 

Students were continuing practicing using the Pythagorean theorem. So I used what I learned from my online class and introduced learning goals and developed success criteria with students using questions and creating buy in with the students. 

I began by drawing a triangle. I labeled one angle with a box denoting it as a right triangle. Then I referenced our investigation of acute obtuse and right triangles. I said that when you have a right triangle the side lengths have special names. They said the diagonal side is the hypotenuse. A few students asked if that's the longest side and other students chipped in that it was. 

Students volunteered the sides to the left and right of the right angle were calls the legs. I decided to review this from Friday because students in second period were solving for the hypotenuse, c, when given c and a leg. So, I made it more explicit in later periods.

To address the first learning goal I developed success criteria with the students that encouraged students to show work at an 8th and 9th grade level.

That began with writing the Pythagorean theorem. Next was substitution. Then simplifying by squaring. Then either adding if solving for the hypotenuse, or subtracting to isolate a variable if solving for a leg. Then square root both sides, to undo the squaring of the variable.

The second learning goal was to identify whether there answer was rational or irrational. I asked students how we could know if the square root of a number was irrational. Most students say it is rational if the square root results in a whole number. We have to be careful here though because the square root of 0.25 is rational because the square root is 0.5, which is a decimal, not a whole number. I moved students to realizing that the radicand must be a perfect square for it to be rational.

Then students practiced on a few problems, as well as interpreting the distance between 1st and 3rd base in baseball when the distance between the bases is 90 feet.
Sample work after developing success criteria.
This student clearly was following the success criteria, showing work superbly.
An interesting problem was 9-124. It gave a triangle on the coordinate plane and you had to find the perimeter. I and most other students saw it as a rectangle with triangles missing out of it, leaving you with 2 pythagorean theorem problems. Amir in 4th period saw it in a much simpler way: draw a horizontal line to split the triangle into 2 triangles. Then use the Pythagorean theorem to solve for the missing side lengths. I liked that elegant method.
After finding the missing sides and perimeter, students were pushed to prove if it was a right triangle. They realized it was acute. This was the first time they saw a square root of a number being squared, which was unfamiliar to them.
In accelerated students identified the difference between intercept and intersect. Two functions intersect when they overlap. A function has an intercept when the function crosses the an axes (x or y). I asked them what those coordinates had in common. They said there was at least 1 zero in the x or y coordinate.

Students used substitution to find points of intersection between quadratics and linear functions. Then Robert showed how once you solved for the roots, these were the x coordinates of the intersection points. He then said to substitute the x values into the linear equation since it would be much easier than substituting into the quadratic.
Find the intersections of functions.
My friend invited me to my first Giants game of the season. We ended up winning, 5-4! Went on too late, so had to leave early.

Sunday, April 24, 2016

Day 144: Pythagorean Theorem Practice & Exponential Equations

Today was a difficult estimation. I was pleasantly surprised at the amazing thinking some of my students produced. I admit that I am not the greatest with geography, and the estimation was the flight distance from Los Angeles, CA to Dulles, VA. Many of my students and I had taken a similar trip when flying to Virginia for the Sojourn to the Past trip so some had more background knowledge than others.

I was impressed with one student who watched the monitor on the back of the seat in front of him during a flight and saw that the cruising speed of the airplane was 300 miles per hour. The flight took 7 hours to the east coast, so he estimated 300*7 or 2100 miles. Wow.

Another great line of thinking was a student who knew that to drive from SF to LA it's about 400 miles. So, that's about 1/7 the trip to the east coast, so he did 400 times 7 and got 2800. Both great lines of thinking.

Today was the followup to Dan Meyer's taco cart. I had different students start off class by summarizing the question we were attempting to answer yesterday, and to briefly describe how we figured out the time it took each person to arrive at the taco cart. It was great to hear them summarize it.

I started by having students draw a right triangle. I asked them what the longest side of the right triangle is called. Some remembered it was called the "hypotenuse." I had them chorally repeat it after me, because they can have difficulty pronouncing it sometimes. They remembered the shorter sides were called the "legs." I reiterated the hypotenuse is always directly across from the 90 degree right angle.

I asked them what the relationship was between the side lengths, and they stated a squared plus b squared equals c squared. I asked if it mattered what order we add numbers in. They said no, so I said therefore it doesn't matter which leg is a or b, they could be either. Then students found the hypotenuse if the legs were 24 and 32. They also found a leg when given one leg and a hypotenuse. Students told me what to write on the board and unfortunately I didn't snap any photos of their work.

I stressed that when solving you start with writing the equation, showing your substitution, simplifying, and showing all steps. While they are still new to the concept, I want them to know the 8th and 9th grade expectation. I reminded them that you can get partial credit if you make a simple calculation error and if it can be easily found in your organized work.

Establishing the vocabulary associated with the theorem.
Modeling showing all your work. Students can get confused at the last step to undo the squaring by square rooting.
Establishing more background knowledge.
In a students mathography I was bummed to see that 4 years ago his 4th grade teacher was still doing timed tests. Bummer.
In accelerated I assigned each group a different classwork problem for them to show their solving steps on a white board. One group had a cube root problem and a couple students pointed out that you don't do plus or minus when cube rooting 8, because -2 cubed is not equal to positive 8. That was a great little discussion. The classwork was a mixture of absolute value, exponential equations, square root equations, and more.
This group showed how to rewrite 36 as 6^2 so that each expression had the same base of 6.
Nicholas showed what he had learned in Kumon to use logs. It looks mathematically correct to me!
Myself and some of my colleagues celebrated a double baby shower for myself and Mrs. Bronson. You can see the firepit in full effect here.
And some rain came, along with a nice rainbow. The canopy worked fairly well.

Thursday, April 21, 2016

Day 143: Dan Meyer's Taco Cart & Imaginary Numbers

I saw this awesome succulent in the sunset. Anyone know what it is called? It was rather large.
Students estimated how many shopping carts were in a long line. Students used the pillars as reference points and sectioned it off.

 This was after I went over exterior angle theorem and why an acute triangle's small and medium side lengths squared add up to greater than the longest side squared. I had students draw an equilateral or equiangular triangle. They reasoned 180/3 is 60 degrees. I then supposed the sides lengths were 3, and squaring them all results in 9, 9, and 9. 9+9=18 and 18>9 so this generalization must be true for all acute triangles. Therefore, obtuse triangles the inequality faces the opposite direction.

Then I extended an exterior angle. Students said it was obtuse, which was true, but I made sure they realized that it could be an acute angle as well. They saw the 2 angles 60 degrees and the exterior angle as a linear pair or supplementary angles. 180-60 resulted in 120. When you add the 2 remote interior angles, 60 and 60, you get 120, the measure of the exterior angle.

Then students wrangled with Dan Meyer's Taco Cart. For each class I recorded their Act 1 questions, notice, and wonderings. Below that I had them focus on who would arrive to the taco cart first. I modified the question in later periods to who would arrive first and by how much time? Here they are:
2nd period
Act 1:

They both want to go to the same place (taco cart), but want to take separate ways. - Lucas

What are we trying to figure out?

The routes they went formed a triangle.

Which route is faster to get to the taco cart?

There is concrete and sand pathways. Sand is hard to walk in. So it’s slower.

I think the other guy’s route sand and then the concrete is longer.

I think it will even out. Sand will go slower. Other guy is on sand, then going faster on the concrete.

Act 2:

How far is it from the taco cart to where they are (in the sand)?

How far is where they are from the concrete?

How far is the taco cart from the concrete?

How fast do they go in the sand? How fast on the concrete?

3rd period

Act 1:

Are they going at the same pace? - Heidi (no)
I notice they made a triangle. - Amir
What is the distance in the sand from where they are to the street and the sidewalk to the taco cart?

What is distance from them to the taco cart? - Mikaela

The person in the sand was going slower. Concrete was normal. - Makenna
When they showed the dotted lines it formed a slope. - Kyle
Did they get there at the same time? (NO) {I don't know why I answered that question}
DId they start at the same time? (yes)

Who got to the taco cart first?

Act 2: What do you want to know?
How long did it take them?

What is the speed on sand?
What is their speed on concrete?
The distance each person went in the sand, and in the sand and concrete?

4th period:
Act 1:

Is it a right triangle? (Yes) - Roxanna
How slowly do you walk in the sand compared to the street? - Louis
I noticed there were 2 plus signs in the sand. - Bing
How far is the taco cart from the start position of Ben and Dan? - Andrew
Are they walking at the same pace? - Natalie (no)
Will weather slow them down? - Jonathan (no)
Can ben and dan run? No, they are walking. (Sami)
What is the pace?
Do they both have the same leg strides ? -Jonas (let’s assume yes)
Which one is the faster pace?
Dan is going to the sidewalk. Then on the concrete forming a right angle.
Ben is going diagonally.
Who gets to the taco cart first?

Act 2:

How fast are each of them going?
What is the distance between the starting point and the taco cart?
What is the distance between the start point and the concrete and the concrete sidewalk and the taco cart?
How much faster do you walk on concrete than on sand?

Period 5:
Act 1:
What is the speed when you walk on the sand and the speed when you walk on the sidewalk? -Eyad
What is the distance from where they are standing to the taco cart? - Ethan
Who got there first? - Joey
How far is it from the sand to the sidewalk? Gracyn
What’s so good about the tacos?
Is it a race of who gets there first?
Are they walking at the same speed?
Is there anything delaying them?

Act 2:
What is the distance from where they are standing and the taco cart?
What is the distance from the sand to the sidewalk?
How fast are they walking?
What is the distance from the taco cart to where he gets off the sand and walks on the concrete?

What is the speed when you walk on the sand and the speed when you walk on the sidewalk?

Some students looked up the pythagorean theorem to use. Some thought to add up the sides and divide by 2. Most solved Dan's route first. The key there was finding the time using the distance and rate using separate rates. More students last year thought the diagonal was the 2 sides added together. This year very few thought that. I did suggest some to take out their data sheets where they investigated acute, obtuse, and right triangles.

Some realized the side lengths needed to be squared to find the area of the squares that form the side lengths. Once they squared both distances and added them together, they reasoned they had to square root that sum to get the 3rd side, or hypotenuse.

Like the 5 Practices book suggest, I tried to get a student that hadn't figured out the diagonal to present how they found Dan's route using their knowledge of rates. Then I had another student present about how they found the diagonal's length and divide it by the rate they walk in the sand.

I think that Estimation 180 helped students be more successful with interpreting the seconds, but students still sometimes interpret decimals as portions of minutes. When I encouraged long division to double check they realized the remainder was the number of seconds.


In this class I had time to show how the diagram related to the pythagorean theorem on the right.
Here you can see the quotient 4.58 minutes. Some students mistook this as 4 minutes 58 seconds. Also, I remember telling students before this, if I go for a walk do I say it took me 275 seconds?
The larger numbers made some of the students' eyes get a little wider.
In accelerated I reviewed the part c of the problem. Students were having trouble simplifying the square root of -81. I modeled that =81=81 *-1 so I can substitute that. Then I can square root 81, followed by square rooting -1, and replacing that with i. Therefore, the answer is plus or minus 9i.

I broke it down into building blocks with the square root of 4 times itself. I simplified each factor to 2, multiplied 2 by 2 to get 4. I reasoned that I also could have multiplied the radicands to get a product of 16, and the square root of 16, is also 4. I don't recall the textbook or myself discussing radicals and simplifying them. (note to self: include that in my paternity sub plans).

Unfortunately I forgot to take a picture of 2 different students who solved quadratics and used "i" in their answers.

Wednesday, April 20, 2016

Day 142: Terminate or repeat? Discriminant. Imaginary Numbers Intro

Today was the last day of estimating song lengths. I was sad. It went out with a marathon song, Jethro Tull's Thick as a Brick. Notice the contrasting reasoning here. The first student multiplied 5:47 by 3 and got 16:41. Another student thought it was 1/4 of the song, but converted to seconds first and then multiplied by 4. This got him 1388 SECONDS, which was then divided by 60 to get 23 remainder 8.

I pointed out to all classes how important it was to interpret the different units of minutes and seconds and asked why the first method didn't work. I also liked how some classes students shared their distributive property thinking and did 3 times 5 minutes, to get 15. Then 3 * 47 then converted the last product to minutes and seconds, then added those together.


Interpreting the division process.
Here's the student with distributive property thinking.

The classwork was continuing investigating terminating versus repeating decimals. I loved how I got as many explanations as I could get for why 57/100 was not equivalent to 0.57 repeating. Not pictured is students saying that one was .57575 and if rounded was .576. The other fraction, 57/100, terminates. It's 0.57. I added that a terminating decimal can have as many zeroes to the right of it and not change the value. One student shared this thinking, then subtracted them. You got a remainder she said, which meant that the two values were not the same.
Terminating decimals. Comparing. Then fraction to repeating decimal. Students reasoned that the number of 9's in the denominator told you how many numbers were repeating. For example, 19/999 is .019019.... because there rare 3 nines in the denominator.
6
I liked this students thinking on the test. 



This student wanted to point out the difference between the thousandths place of each decimal form.
These students called me over. The bottom right calculator gave the number 1/999 as 1.001 x 10^-1. I thought that that was somewhat odd, and discussed it with them.
Students in accelerated used the discriminant to identify how many solutions an equation had without solving it. If b squared minus 4ac was less than 0, there were no real solutions. If it equalled zero, one solution. If it was greater than 0, it had 2 solutions. Then students were introduced to imaginary numbers. Tomorrow we will jump off with 0=x^2+81, and I want to illustrate to students that the square root of 4 times the square root 4 is 4 for 2 reasoned. The square root of 4 is 2, so 2*2=4. If you multiply the radicands, 4 and 4, you get 16. The square root of 16 is 4, proving that when you multiply radical numbers, you multiply the radicands by each other.

When the text said there were 2 solutions, some students realized that when you square root both sides you can't forget the plus or minus sign. This equation has positive and i and negative i as a solution. 2 imaginary solutions.

Day 141: Rational, Irrational Numbers & Proving the Quadratic Equation

Students estimated Jimi Hendrix All Along the Watchtower, day 133 of estimation 180. I didn't take a picture of sample reasoning.

I read the introduction of the section to students detailing how rational numbers can be written as a ratio, and since 1/3 is a repeating decimal, it is also concerned a rational number. I taught students how to square root a number using the TI 83 calculator and I also had them take out their iPhones or Androids to use the square root button. On the Android you have to press the radical or square root button first before the radicand. On an iPhone you can do it either way.

We also discussed how 5/9 showed up as .555555556 and I asked students why that was and what it meant. They said it was repeating and the calculator was rounding the last digit.

For some reason students questioned if the negative sign outside the radical affected it being rational or irrational. Some students pressed the negative sign before the square root button and there calculators gave them an error. One student knew that it was impossible to square root a negative number.

Some students mistook pi squared as pi times 2, and students were quick to correct that 2 was an exponent so it was pi times pi. I was relieved that a few students remembered scientific notation with the last expression thrown in. Some students reasoned you move the decimal to the left twice. Others said that 10^-2 is like .01 or 1 one hundredth so they multiplied that by the decimal.

This also re-introduced fractions with 9's in the denominator as repeating in a certain pattern.

Students summarized that when you square root a non-perfect square, the result goes on forever in a non-repeating pattern. Therefore the square root of 9 is rational because it is equal to 3 (3/1). 
We reviewed the fact that consecutive whole numbers when summed are equal to the larger number squared minus the smaller number squared. Students explained how I should foil, and then cancel out the x squared terms, leaving the 2 expressions equivalent. I demonstrated and reinforced the generic rectangle for multiplying (x+1) squared.
A student in 6th period was investigating growth of quadratics and the relationship to derivatives and slope.
In accelerated students copied the steps to solving the quadratic formula from the standard form of the quadratic equation. The biggest road block was the 3rd step of completing the square. Owen said that when you complete the square you add c to both sides, which in a perfect square trinomial is (b/2)^2. In the quadratic equation step, b is b/2, so students have to make the jump to see b/a divided by 2 is b/2 multiplied by 1/2, which is b/2a. Then it is (b/2a)^2. Which equal b^2/4a^2 added to both sides of the equation.

CPM has students complete a generic rectangle for this step which they started with x^2 in the bottom left corner. The top right corner is b^2/4a^2 and the other diagonal is the middle terms in a perfect square trinomial: (b/2a)+(b/2a).

Then you write it as a perfect square trinomial. I would personally reverse steps 4 and 5. I think of completing the square as adding a term to both sides. I didn't take a picture of the rest of the steps, but next is commutative property, followierd by making c/a a like fraction by multiplying it by 4a, leaving you with b^2-4ac/4a^2. When you square root both sides the plus or minus sign is introduced and the denominator simplifies to 2a. The numerator is the opposite of b plus or minus the square root of b squared minus 4ac. We discussed yesterday how this radicand, the discriminant, determines if there's no real solutions, one solution, or two solutions.
Wrestling with proving the quadratic equation.

I want to make a sign out of math is supposed to make sense and "in this class mistakes are expected, inspected, and respected!"
I liked this answer to D for a test question, but apparently the percentage of students wearing a backpack is highest of the whole school.
2 way table assessment question

Monday, April 18, 2016

Day 140: Approximating Irrationals & Discriminant Intro

Today's estimation was the Star Wares main title song. Some students elected to not use the clue and "go off experience." Those that thought 1/5 of the song had been played at 47 seconds got a close estimate. Once again I asked students to tell me how they interpreted the seconds as minutes and seconds. These discussions dig deep into place value understanding.

Students had background knowledge via notes on perfect squares, and a procedure for approximating an irrational number to the nearest tenth which brought up some amazing discoveries in 4th period. Read on below for that.

I also introduce a learning goal, and unfortunately did not develop the success criteria with students. The goal was to approximate, without a calculator, an irrational number to the nearest tenth. Another learning goal that I should have written was interpret the square root of a perfect square as well. We also did a participation quiz in each class to reinforce study team norms.
The number line method has students find the 2 perfect squares less than and greater than the radicand of the irrational number.
A student at table 5 was convincing his group that the number was between 4 and 5, and not 4 and 6. He said 4 squared was 16 and 5 squared was 25. Group 8 used notes, and one member used a calculator to approximate first. 1 team member continued to see if the group was following along. For Table 7 they had different answers so recorder reporter wasn't making sure students wrote it down.
Here a student successfully approximated the square root of 95. They put the square root of 81 and 100 on opposite sides of a number line and placed root 95 closer to the right. It was 81 and 95 are 14 spaces away, and the 2 perfect squares are 19 away, making it 9 14/19. This student did not convert it to decimal form after that. Interestingly, 100-81 is equal to 10+9.
Here is the example I did when directed by a student volunteer. the radicand of 40 is between the 2 perfect squares 36 and 49. They are 13 apart, while the square root of 40 is 4 away from the perfect square ON THE LEFT, meaning the square root of 40 is approximately 6 and 4/13. I'd like to make a desmos activity that scaffolded how to do this with students.
Table 5 had a group member that was difficult to deal with for the group, the class, and myself. We reinforced the study team norms. Group 6 asked me a question when the whole group had a try at answering it. In group 4 a student was working on the problem while his team mate was reading it. Group 1 stated that 24 is not a perfect square, which is why you couldn't make a square out of 24 square tiles.


A joke a student made up and showed you. "Four root punch." Get it?
Group 5 referred to their knows, group 6 checked to see what everyone got for the side lengths of perfect squares. Table 2 explained the number line method from the notes to a team member.
Here is where students noticed 49-36, is equal to +7.
This is my first attempt at wording it. I told a colleague at lunch and she found this resource, where I like the intro to multiplying binomials. There is a NEED to prove this:
My colleague Carrie found this resource for what Bing noticed. http://www.transum.org/Software/SW/Starter_of_the_day/starter_October10.ASP. The dub it "Consecutive squares."
I tried to elaborate here.
I liked how group 5 checked on if others were going ahead, and the one who went ahead said they didn't want to hear about her drama. Nice job standing up! One group thought the square root of 49 was written as 7^2, and realized it couldn't. Group 4 had 2 members interpreting the notes. They got the idea, except the larger perfect square wasn't the closest one to that number.
Showing dividing 4 by 13. A student predicted you'd get a more accurate decimal if you made the mixed number an improper fraction. We checked this, and saw it gave the same decimal approximation.
In accelerated we reviewed quadratics in perfect square form (ax-b)^2=c^2. They realized if c was negative, you'd get no real solutions. If it was zero you'd get 1 solution. If it was greater than zero you got 2 solutions.

We expanded this to standard form when I asked student to solve 2x^2-3x+1/2 using the quadratic formula. I gave them time, I then had a student recite me the quadratic formula, and I stressed that it was "all over 2a" and if you forgot to underline the opposite of b it was completely wrong. I also stressed where I saw students make the most frequent mistakes and it was at -4ac. We talked about exact and approximate decimal form.

Basically the radicand, b^2-4ac, determines how many solutions there are students saw.

The lesson asked students to come up with a c value that made the equation have zero solutions, and for the equation to have no solutions. Students realized there'd be no solution if the radicand had the variable c and was less than 0. So, when solving that inequality, if c was greater than 9/8, there would be no real solutions. If it was 9/8, there would be one solution. Finally, if it was less than 9/8, there would be 2 solutions.

I asked students who took Kumon if they knew what b squared minus 4ac was. Moreen remembered it was called the discriminant. Therefore I asked students what was true about it for each situation. They said if it equals zero there's one solution, if it's less than 0, no real solutions, and if it's greater than 0 there are 2 solutions.
Investigating quadratics in depth. This makes me want to connect it to graphs with the Desmos activity I saw done by Shelly Caranza.
Annabel investigated the conjecture made by a student in an earlier class. Love how she rewrote the squaring as repeated multiplication and then as repeated addition to show how the difference between the numbers was 1, leaving you with 1 of the larger number left over, plus 1.
Nuri reviewed how to solve this absolute value equation. The extension was what would make this equation have no solution. They said if the expression on the right was a negative number. Basically, the absolute value of an expression will equivalent to 9 or -9 when the expression on the right is a positive 9.