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.
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.

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