January 3rd, 2017, 01:24 PM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,146 Thanks: 685  Squaring twice
I presume this is about the simplest equation where solving for x requires "squaring both sides" twice: sqrt(x) = sqrt(ux)  v A few gyrations leads to: x = [v^2 (2SQRT(u) + u + 1)] / (u^2  2u + 1) Is that something "known"; like, used by students as a short cut to these headachy equations? 
January 3rd, 2017, 01:29 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,302 Thanks: 666 
did you mean to reply to some thread w/this?

January 3rd, 2017, 02:11 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,734 Thanks: 1360 
If u = 1, v = 0 and x can have any value. If u isn't 1 and both x and u are nonnegative, v = √(ux)  √x = (√u  1)√x. So far, no squaring. 
January 3rd, 2017, 03:46 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,146 Thanks: 685 
Ya'll missing my point. I mean: when squaring both sides IS REQUIRED in order to solve. I was thinking along the lines of: students are aware that (a+b)(ab) = a^2  b^2; so there is no need to do the "long multiplication". I was wondering if there is a similar short cut that (at one point) students are given, in order to shorten all the work involved in squaring both sides. If I'm still not clear, forget it! 
January 3rd, 2017, 04:21 PM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,290 Thanks: 441 Math Focus: Yet to find out. 
Won't there be too many situations when this happens though? I don't see the short cut part seems like the same amount of effort to rearrange then remember the cutting short.

January 4th, 2017, 03:46 AM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 
How about you do this $ \sqrt{x}( \sqrt{u}  1) = v $ $ x = ( \frac{v}{\sqrt{u}  1} )^2 $ Since $ u , v $ are constants I presume , stop here then 'plug and play' 
January 4th, 2017, 03:59 AM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,146 Thanks: 685 
But that's a completely different case, Agent. Same as sqrt(x) = c (since u and v are constants/givens.) So x = c^2 
January 4th, 2017, 04:09 AM  #8 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 
Yes but your 'gyrations' have not eliminated the square root symbol either and are more lengthy ...

January 4th, 2017, 04:14 AM  #9 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 
You may be thinking of 'harder' things that make squaring twice inevitable for example ... $\sqrt{x} = \sqrt{u + x}  v $ Denis? 
January 4th, 2017, 04:30 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,146 Thanks: 685 
Yes, I've already stated that twice: cases where the ONLY way to solve is squaring both sides twice. 

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