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Expert replies
by voodoo_child » Mon Apr 30, 2012 6:40 am
Find the solution of Sqrt(2+sqrt(2+sqrt(2+sqrt(.........

Here's what I did:

sqrt(2+x)=x
Therefore, 2+x=x^2
Therefore, x^2 - x - 2 => (x-2) (x+1)=0
X = 2 or -1.

Correct answer is 2; I am not able to rule out (-1). Any help? Thanks
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Source: — Problem Solving |

by aneesh.kg » Mon Apr 30, 2012 6:43 am
Square root of any quantity is always positive.
Since x is square root of a quantity here, the negative value (-1) is ruled out.
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by Brent@GMATPrepNow » Mon Apr 30, 2012 8:00 am
voodoo_child wrote:Find the solution of Sqrt(2+sqrt(2+sqrt(2+sqrt(.........

Here's what I did:

sqrt(2+x)=x
Therefore, 2+x=x^2
Therefore, x^2 - x - 2 => (x-2) (x+1)=0
X = 2 or -1.
Correct answer is 2; I am not able to rule out (-1). Any help? Thanks
When you solve a question involving square roots, you should plug in the answers to check for extraneous roots (aka, solutions that don't work).

x=2: sqrt(2+2)=2 (WORKS)
x=-1: sqrt(2+(-1))=-1 (DOESN'T WORK)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by voodoo_child » Mon Apr 30, 2012 11:06 am
Brent@GMATPrepNow wrote: x=2: sqrt(2+2)=2 (WORKS)
x=-1: sqrt(2+(-1))=-1 (DOESN'T WORK)

Cheers,
Brent
Brent and Aneesh - question - Sqrt (1) is not equal to -1? Why? (-1)^2 =1. Therefore, Sqrt(1) = 1 or -1. Correct?
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by Brent@GMATPrepNow » Mon Apr 30, 2012 11:13 am
voodoo_child wrote:
Brent@GMATPrepNow wrote: x=2: sqrt(2+2)=2 (WORKS)
x=-1: sqrt(2+(-1))=-1 (DOESN'T WORK)

Cheers,
Brent
Brent and Aneesh - question - Sqrt (1) is not equal to -1? Why? (-1)^2 =1. Therefore, Sqrt(1) = 1 or -1. Correct?
By definition, sqrt(n) denotes the positive number whose square is n.
Similarly, -sqrt(n) denotes the negative number whose square is n.

So, the takeaway here is that the equation x^2 = 9 has 2 solutions (x = 3, -3), while the equation sqrt(9)= x has only 1 solution (x = 3)

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by aneesh.kg » Mon Apr 30, 2012 11:44 am
By definition, when the symbol of 'square root' is placed over a quantity, the output must be positive. Notice that you don't get two values when you take the square root of a number on your calculator.

To clear your confusion, let's take this example:

say x^2 = 9
Step1: x = +/- sqrt(9)
Step2: Since sqrt(9) = 3,
Step3: x = +/- 3

So,
(i) if x^2 = any number, we get two values - one positive and one negative - of x
(ii) if x = sqrt(number), we get just one positive value of x
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