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Intern
Joined: 01 Dec 2012
Posts: 1

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01 Dec 2012, 14:41
Can anyone explain how to solve a problem like this one?
sqrt(y+(sqrt(y+(sqrt(y)))))=2

What is the value of y? That's the entire problem.

Last edited by A777 on 02 Dec 2012, 09:49, edited 3 times in total.

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Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622

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Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)

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02 Dec 2012, 00:55
A777 wrote:
Can anyone explain how to solve a problem like this one?

sqrt(y+2(sqrt(y+2(sqrt(y)))))=

Can you post the full problem?
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How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

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Intern
Joined: 27 Nov 2012
Posts: 37

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02 Dec 2012, 14:06
Square both sides. Keep doing this until there are no square roots.

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Manager
Joined: 02 May 2012
Posts: 108

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Location: United Kingdom
WE: Account Management (Other)

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04 Dec 2012, 07:50
The equation can be rewritten as 2 = Sqrt(y + sqrt(y +.....))
So square both sides:
4 = (y + sqrt(y +.....))

So if 4 = (y + sqrt(y +.....)) and the original equation has an infinite number of (y + sqrt(y +.....)) terms, then you can substitute x^2 in the original equation.

I.e.
2 = Sqrt(y + sqrt(y +.....))
2 = Sqrt(y + 4)
+/-2 = y + 4

So y = 2-4 = -2
Or y = -2-4 = -6
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Kudos [?]: 74 [0], given: 34

Manager
Joined: 02 May 2012
Posts: 108

Kudos [?]: 74 [0], given: 34

Location: United Kingdom
WE: Account Management (Other)

### Show Tags

04 Dec 2012, 07:50
The equation can be rewritten as 2 = Sqrt(y + sqrt(y +.....))
So square both sides:
4 = (y + sqrt(y +.....))

So if 4 = (y + sqrt(y +.....)) and the original equation has an infinite number of (y + sqrt(y +.....)) terms, then you can substitute x^2 in the original equation.

I.e.
2 = Sqrt(y + sqrt(y +.....))
2 = Sqrt(y + 4)
+/-2 = y + 4

So y = 2-4 = -2
Or y = -2-4 = -6
_________________

In the study cave!

Kudos [?]: 74 [0], given: 34

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