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Old 2007-06-18, 22:36   #1
davar55
 
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Default Solve this equation

Solve for x:

(x+9)^(1/3) - (x-9)^(1/3) = 3

(This may be too easy for this forum,
but I've always liked the solution method.)
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Old 2007-06-18, 23:23   #2
ewmayer
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x[sup]2[/sup]=80

Last fiddled with by ewmayer on 2007-06-18 at 23:23
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Old 2007-06-19, 07:50   #3
davieddy
 
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Quote:
Originally Posted by ewmayer View Post
x[sup]2[/sup]=80
Just as well this wasn't the homework help forum
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Old 2007-06-19, 12:20   #4
mfgoode
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Question Solution!

Quote:
Originally Posted by davar55 View Post
Solve for x:

(x+9)^(1/3) - (x-9)^(1/3) = 3

(This may be too easy for this forum,
but I've always liked the solution method.)


Lets have your method davar. Thank you!

Mally
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Old 2007-06-19, 12:56   #5
Kees
 
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Taking cubes on both sides gives the result without any noteworthy manipulation.
If there is an elegant solution or a smart substitution it still needs to be very short to beat the straightforward one

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Old 2007-06-19, 14:23   #6
VolMike
 
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Equation doesn't have solution(s)
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Old 2007-06-19, 15:40   #7
davieddy
 
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Quote:
Originally Posted by Kees View Post
Taking cubes on both sides gives the result without any noteworthy manipulation.
The entertaining stage is:

-3[(x+9)1/3-(x-9)1/3](x2-81)1/3 = 9

at which point we use the original equation to substitute 3
for [...]

David

Last fiddled with by davieddy on 2007-06-19 at 15:42
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Old 2007-06-19, 15:51   #8
Mini-Geek
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Quote:
Originally Posted by VolMike View Post
Equation doesn't have solution(s)
x=sqrt{80}
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Old 2007-06-19, 21:14   #9
fetofs
 
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Not quite.

x = \pm\sqrt{80}
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Old 2007-06-20, 07:32   #10
mfgoode
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Lightbulb Solutions!

Quote:
Originally Posted by davar55 View Post
Solve for x:

(x+9)^(1/3) - (x-9)^(1/3) = 3

(This may be too easy for this forum,
but I've always liked the solution method.)


A good problem but not instructional enough.

Its not good nor instructional getting straight solutions from the 'know all's'.

One right answer as ewmayer's is enough to tackle the problem

A hint from Keyes was good as it saved a lot of 'computation'. I have to use the term in these computer days as this is piffle for the comp.

As usual Davieddy's 'solution' and as he termed it 'entertainment' made things more confused and you can all see for your selves. Imagine substituting a whole equation for an integer ? And he claims to be a math's tutor !

A little knowledge is a very dangerous thing. Arrogance is even worse!

Now given that the solution is correct (I have not checked it out but accept Ewmayer's as correct) the next question is to generalise it from a particular problem.

My question is how would you solve the problem when the exponents are not equal? Lets say they are p and q and not just 1/3 ? How would you get the rationalising factor esp.
when there are three or more surds?

Thank you Davar55 as it made me revise my algebra!

BTW: if you have a better solution than keyes kindly respond and let me know.

Mally

Last fiddled with by mfgoode on 2007-06-20 at 07:35
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Old 2007-06-20, 08:03   #11
Chris Card
 
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Quote:
Originally Posted by mfgoode View Post


As usual Davieddy's 'solution' and as he termed it 'entertainment' made things more confused and you can all see for your selves. Imagine substituting a whole equation for an integer ? And he claims to be a math's tutor !
What are you on about Mally? Davieddy's solution is exactly the way to do it. The substitution he suggests is valid, and the simplest way I can see to get to the solution.

Chris
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