The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Professional Associations » nyshsmath

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: re: #32
Replies: 23   Last Post: Jun 18, 2010 5:18 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 102
Registered: 8/26/09
Re: #32
Posted: Jun 16, 2010 4:38 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply
att1.html (2.8 K)

hmmm.. actually what if x = -2? Then sq rt of (-2)cubed = 2i sq rt 2. But if x = -2 and you are doing the sq rt of (-2) squared that would still just be 2. I think. In which case you are right. I think. I'm rethinking all my math knowledge now!!
----- Original Message -----
Sent: Wednesday, June 16, 2010 3:45 PM
Subject: re: #32

Which is why is should be absolute value of x, because while you assume the positive square root (principal root), you don't know that x is positive - it was never indicated that the domain was positive values of x.
(I'm not sure if we're in agreement, or doing two different things.)

when solving an equation x squared = 9, then yes the answer is + and - 3, but when it is written as a square root, you assume the positive square root.

******************************************************************* * To unsubscribe from this mailing list, email the message * "unsubscribe nyshsmath" to * * Read prior posts and download attachments from the web archives at * *******************************************************************


No virus found in this incoming message.
Checked by AVG -
Version: 8.5.437 / Virus Database: 271.1.1/2940 - Release Date: 06/15/10 18:35:00

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.