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Is x = \sqrt{x^2} 1. x = even 2. 13<x<17

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

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New post 21 Mar 2007, 09:07
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Is \(x = \sqrt{x^2}\)

(1) x = even
(2) 13<x<17

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-square-root-of-x-126060.html
[Reveal] Spoiler: OA
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New post 21 Mar 2007, 14:33
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(B) for me :)

We should transform a little the question to know clearly what it is asked :)

x = sqrt (x^2) ?
<=> x = |x| ?
<=> x >= 0 ?

Stat 1
x is even
<=> x = 2*k where k is an integer.

So,
o If k > 0, then x > 0
o If k < 0, then x < 0

INSUFF.

Stat 2
13 < x < 17
=> x > 0

SUFF.
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New post 22 Mar 2007, 05:22
Hi Summer, thanks for the link but frankly speaking I could not get amardeep's explanation for statement 2..may be I am bit dumb or more dumb :cry:

Can anybody please explain the same...thanks
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New post 22 Mar 2007, 05:37
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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.

:)
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New post 18 Jul 2007, 12:23
Sorry Fig.. I didnt get ur approach :?

As i see it

For a number 14 : sqrt(14^2) would give both +14 and -14 as the answer.

Hence x != sqrt(x^2)

Maybe my concepts are screwed up.. Help
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Re: Is x = \sqrt{x^2} 1. x = even 2. 13<x<17 [#permalink]

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New post 16 Jun 2013, 00:57
I thought, even numbers had to be positive but as I learnt, they can be both +ve as well as -ve.
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Re: Is x = \sqrt{x^2} 1. x = even 2. 13<x<17 [#permalink]

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New post 16 Jun 2013, 01:15
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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).

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Is \(x = \sqrt{x^2}\)?

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.

Answer: B.

Similar questions about this concept:
is-root-x-3-2-3-x-92204.html
if-x-0-then-root-x-x-is-81600.html
is-sqrt-x-5-2-5-x-100517.html
if-x-0-then-root-x-x-is-100303.html

Hope it helps.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-square-root-of-x-126060.html
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Re: Is x = \sqrt{x^2} 1. x = even 2. 13<x<17   [#permalink] 16 Jun 2013, 01:15
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