## 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.*

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 amWell 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 pmColleen – 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 amCase 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 pmPretty clean, David! Thanks for sharing.

5.David Radcliffe (@daveinstpaul) | September 19, 2011 at 11:12 amOops. Please replace 18 with √18 at the end of the last line.

6.eadurkin | September 19, 2011 at 4:33 pmI 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 amThanks for sharing! This is pretty efficient.

8.Stefras (Shawn Urban) | September 19, 2011 at 10:36 pmI 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

xuntil 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 amShawn – 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[…] was pretty interesting. I posted a blog asking for solutions to a problem, and got a lot of […]