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There is a tricky aspect to remember about sqrt(x)... The function sqrt(x) is only definied for x >=0 and the function returns only one value for a positive or nul x.

That said... This is the trick:
> x^2 = b^2 : x can be either b or -b
> sqrt(Y) = c : y can only be c^2 (no -c^2) >>> y must be positive by definition of sqrt()

So... the answer is (B) not (E)... and actually, this question is just asking x >= 0.

Re: Is x = \sqrt{x^2} 1. x = even 2. 13<x<17 [#permalink]

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16 Jun 2013, 01:15

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This post received KUDOS

Expert's post

avik629 wrote:

I thought, even numbers had to be positive but as I learnt, they can be both +ve as well as -ve.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, positive as well as negative numbers could be even. For example, -8, -6, -2 are all even numbers.

Also, note that 0 is also an even integer: zero is evenly divisible by 2 thus it must be even (in fact zero is divisible by every integer except zero itself).

Note that: \(\sqrt{x^2}=|x|\), so the question basically asks whether \(x=|x|\) or, which is the same, whether \(x\geq{0}\) or whether \(x\) is a non-negative number.

(1) x = even --> not sufficient as x can be negative as well as non-negative even number. (2) 13<x<17 --> x is non-negative. Sufficient.

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