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January 3rd, 2017, 02:24 PM   #1
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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?
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January 3rd, 2017, 02:29 PM   #2
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did you mean to reply to some thread w/this?
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January 3rd, 2017, 03:11 PM   #3
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If u = 1, v = 0 and x can have any value.

If u isn't 1 and both x and u are non-negative, v = √(ux) - √x = (√u - 1)√x.

So far, no squaring.
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January 3rd, 2017, 04:46 PM   #4
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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)(a-b) = 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!
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January 3rd, 2017, 05:21 PM   #5
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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.
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January 4th, 2017, 04:46 AM   #6
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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'

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January 4th, 2017, 04:59 AM   #7
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But that's a completely different case, Agent.
Same as sqrt(x) = c (since u and v are constants/givens.)
So x = c^2
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January 4th, 2017, 05:09 AM   #8
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Yes but your 'gyrations' have not eliminated the square root symbol either and are more lengthy ...
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January 4th, 2017, 05:14 AM   #9
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You may be thinking of 'harder' things that make squaring twice inevitable for example ...

$\sqrt{x} = \sqrt{u + x} - v $

Denis?
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January 4th, 2017, 05:30 AM   #10
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Yes, I've already stated that twice:
cases where the ONLY way to solve is squaring both sides twice.
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