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Solve this root

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Manager
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Joined: 25 May 2009
Posts: 142

Kudos [?]: 149 [0], given: 2

Concentration: Finance
GMAT Date: 12-16-2011
Solve this root [#permalink]

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New post 01 Jul 2009, 11:08
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Solve : \((sqrt(n+1))^2\)

Can someone help me solve this root? Please show work. Thanks.

Kudos [?]: 149 [0], given: 2

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Joined: 16 Mar 2009
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Kudos [?]: 996 [0], given: 19

Location: Bologna, Italy
Re: Solve this root [#permalink]

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New post 01 Jul 2009, 16:25

Kudos [?]: 996 [0], given: 19

Manager
Manager
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Joined: 25 May 2009
Posts: 142

Kudos [?]: 149 [0], given: 2

Concentration: Finance
GMAT Date: 12-16-2011
Re: Solve this root [#permalink]

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New post 02 Jul 2009, 05:07
Could someone show work please?

Kudos [?]: 149 [0], given: 2

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Joined: 16 Mar 2009
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Kudos [?]: 996 [0], given: 19

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Re: Solve this root [#permalink]

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New post 03 Jul 2009, 07:30
I3igDmsu wrote:
Could someone show work please?


sorry I didn't have any middle steps ... it kind of popped out of my head...

\((\sqrt{n+1})^2 = n+1,\) where n+1<0, has no solution, for real numbers (cuz \(\sqrt{n} = n^{1/2},\) where n is a positive real number... so \(((n+1)^{1/2})^2 = (n+1)^{{1/2*2}} =n+1\))
BUT
\((\sqrt{n+1})^2 = |n+1|\) for complex numbers.


I guess you would need someone with degree in math to provide more insight to why it is like that, I'm not that good ))
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Kudos [?]: 996 [0], given: 19

Manager
Manager
avatar
Joined: 25 May 2009
Posts: 142

Kudos [?]: 149 [0], given: 2

Concentration: Finance
GMAT Date: 12-16-2011
Re: Solve this root [#permalink]

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New post 05 Jul 2009, 20:08
I was looking to simplify the equation above, rather than solve. Can you use the FOIL method here? I guess what I am getting at is if you have an expression under a root and that expression is squared, then do you FOIL out the two expressions? How do you FOIL with roots?

Kudos [?]: 149 [0], given: 2

GMAT Club team member
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Posts: 115

Kudos [?]: 996 [0], given: 19

Location: Bologna, Italy
Re: Solve this root [#permalink]

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New post 06 Jul 2009, 05:16
I3igDmsu wrote:
Can you use the FOIL method here? I guess what I am getting at is if you have an expression under a root and that expression is squared. Do you FOIL out the two expressions? How do you do it with roots?


You are not making fun of me, are you? :shock:


I will suppose that not and will respond. You mean put it this way \((\sqrt{n+1})(\sqrt{n+1})\)?
No you can't use FOIL because both terms are under one root. You could have used FOIL if it would have been \((\sqrt{n}+1)^2 =(\sqrt{n}+1)(\sqrt{n}+1)\) , but even in that case it would be more logical to use perfect square binomial identity - \((a+b)^2 = a^2+2ab+b^2\) (Note: you really have to know basic polynomial identities. Make sure you know the answer to these - \((a \pm b)^2 = ? a^2-b^2=?\) and maybe \(a^3 \pm b^3=? (a \pm b)^3 =?\))

BUT in your case, as I said \((\sqrt{n+1})^2 = ((n+1)^{1/2})^2\) Note the brackets. You can't use FOIL because of the priority of the operations first you have too take the root (put to the 1/2 power) then square (multiply the brackets)....
To tickle your mind (or maybe to confuse you, so may disregard this part) I could abstractly show this in terms of a function of higher order... like \(f(x) = n+1, g(x) = \sqrt{x}, g(f(x)) = \sqrt{f(x)};\) agh never mind I will make it overly complicated.. I tried to show that while solving smth it really helps me to make an abstraction i.e. treat some polynomial as one unknown variable to see what operations I have to do with it...

@ramiy make sure you don't confuse this \((\sqrt{n+1})^2\) with \(\sqrt{(n+1)^2}\)
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Re: Solve this root   [#permalink] 06 Jul 2009, 05:16
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