Math Teachers – Solve This (#1)
September 19, 2011 at 10:44 am 10 comments
Here’s the first of what I hope will be a running feature on this blog. The premise: I’ll give a problem and ask you to explain how you would show your students how to do it. (Students – I’d love to hear your ideas, too.) The idea came up when a colleague showed me a problem his students were working on, and I started thinking about alternate ways I would try the problem. I came up with a couple of ideas that I thought were more efficient.
The Problem
Solve. Present your solutions using interval notation.
Your Solution
Leave your solution as a comment on this page, or you can send it to me through the contact page on my website.
Edit: I will share my thoughts on Friday the 23rd.
- George
I am a math instructor at College of the Sequoias in Visalia, CA. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com.
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Entry filed under: Math. Tags: absolute value inequalities, algebra, developmental math, george woodbury, intermediate algebra, Math, math teachers, solve this, teaching, woodbury.
1.
Colleen Young | September 19, 2011 at 10:57 am
Well the first thing I would do is draw a picture!
http://bit.ly/p2Pn5D
I love that Desmos graphing calculator – see
http://colleenyoung.wordpress.com/2011/09/17/desmos-graphing-calculator-2/
2.
georgewoodbury | September 19, 2011 at 4:07 pm
Colleen – I like the approach. One strategy I really like is using a graphical approach to get things started. I’ll share mine before the week is out.
Thanks for sharing.
3.
David Radcliffe (@daveinstpaul) | September 19, 2011 at 11:01 am
Case 1: x^2 − 9 ≥ 0
2 ≤ x^2 − 9 < 9
11 ≤ x^2 < 18
Case 2: x^2 − 9 < 0
2 ≤ −(x^2 − 9) < 9
2 ≤ −x^2 + 9 < 9
−7 ≤ −x^2 < 0
0 < x^2 ≤ 7
Solution:
0 < x^2 ≤ 7 or 11 ≤ x^2 < 18
(−√18, −√11] ∪ [−√7, 0) ∪ (0, √7] ∪ [√11, 18)
4.
georgewoodbury | September 19, 2011 at 4:08 pm
Pretty clean, David! Thanks for sharing.
5.
David Radcliffe (@daveinstpaul) | September 19, 2011 at 11:12 am
Oops. Please replace 18 with √18 at the end of the last line.
6.
eadurkin | September 19, 2011 at 4:33 pm
I would do something very similar to David’s solution, but for case 2 I would have the negative signs with the 2 and 9, not with the expression. That is, I would start with the definition of absolute value in terms of position on the number line – a diagram of where the expression x^2 – 9 can lie on the number line would also be a good starting point. So my two compound inequalities would have the expression on the intervals [2,9) and (-9,-2]. Then solve algebraically, same as David’s case 1, but slightly differently to his case 2 as the negative would not be in the middle part of the inequality.
7.
georgewoodbury | September 20, 2011 at 5:38 am
Thanks for sharing! This is pretty efficient.
8.
Stefras (Shawn Urban) | September 19, 2011 at 10:36 pm
I too use a diagram, similar to those used by Kate Nowak in her post, Algebra 2: Solving Absolute Value Equations.
I ignore the squaring of x until after I draw the diagram, which shows (0,7]U[11,18). Then I consider that square-evolution (square-rooting) of the relation will:
1) create a mirror of the original diagram in the negative X-axis
2) square-root the terms, creating 0, root-7, root-11, root-18
and the negatives of these.
This provides the answer,
(-root-18,-root-11]U[-root-7,0)U(0,root-7]U[root-11,root-18).
9.
georgewoodbury | September 20, 2011 at 5:40 am
Shawn – This is a pretty unique solution, and one that students may conceptually understand and remember. Thanks for sharing!
10. Solve This #1 – Wrap Up « George Woodbury’s Blogarithm | September 23, 2011 at 8:53 am
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