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M28-48

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M28-48  [#permalink]

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New post 16 Sep 2014, 01:44
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Re M28-48  [#permalink]

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New post 16 Sep 2014, 01:44
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Re M28-48  [#permalink]

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New post 21 Oct 2015, 08:26
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I think this is a high-quality question and I agree with explanation. This and M04-07 are duplicates
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Re: M28-48  [#permalink]

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New post 28 Oct 2015, 01:54
How can a negative number be prime? Statement II should limit value to x=1 for prime number 2, I think.. please explain
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New post 28 Oct 2015, 06:16
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Re: M28-48  [#permalink]

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New post 17 Apr 2017, 07:02
what about the rule : 'When the GMAT provides square root sign for an even root,such as x^(1/2) or x^(1/4), then the only accepted answer is the positive root''?
I am asking this because i used this rule to find only one solution (1) in the second statement. Is the rule only talking about x^(even power) ?
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New post 17 Apr 2017, 07:23
alexlovesgmat wrote:
what about the rule : 'When the GMAT provides square root sign for an even root,such as x^(1/2) or x^(1/4), then the only accepted answer is the positive root''?
I am asking this because i used this rule to find only one solution (1) in the second statement. Is the rule only talking about x^(even power) ?
thanks


This rule is not violated. I tried to explain this here: https://gmatclub.com/forum/m28-184547.html#p1593296
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Re M28-48  [#permalink]

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New post 14 Jun 2017, 09:29
I think this is a high-quality question and I don't agree with the explanation. Since a prime number can't be negative, why we are considering x = -1? this should always be x =1. isn't it?
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New post 14 Jun 2017, 09:36
kutput wrote:
I think this is a high-quality question and I don't agree with the explanation. Since a prime number can't be negative, why we are considering x = -1? this should always be x =1. isn't it?


Yes, only positive integers can be primes but you get positive result with both 1 and -1. Please re-read the solution again, plug both 1 and -1 and read this reply: https://gmatclub.com/forum/m28-184547.html#p1593296
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Re: M28-48  [#permalink]

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New post 09 Jul 2018, 17:04
Bunuel



Since \sqrt{x} = lxl i.e. x can be +ve as well as -ve, can you please make me understand why \sqrt{(-1) (-1)} is not equals to (-1). We can take one negative value out of root, and multiple of 2*(-1) would be negative i.e. non-prime number.

Why we have to calculate \sqrt{(-1) (-1)} before taking value out of the root? I know this way it would be 2 and a prime number.
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New post 09 Jul 2018, 20:28
Avinash2303 wrote:
Bunuel



Since \sqrt{x} = lxl i.e. x can be +ve as well as -ve, can you please make me understand why \sqrt{(-1) (-1)} is not equals to (-1). We can take one negative value out of root, and multiple of 2*(-1) would be negative i.e. non-prime number.

Why we have to calculate \sqrt{(-1) (-1)} before taking value out of the root? I know this way it would be 2 and a prime number.


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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New post 08 Jan 2019, 07:08
why is it wrong to simplify the second statement to 2*x since the sq.root and the square cancel each other? If i do that then for the second statement to be true, only when x=1 will the result be a prime. :(
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New post 08 Jan 2019, 07:40
ameralhajj wrote:
why is it wrong to simplify the second statement to 2*x since the sq.root and the square cancel each other? If i do that then for the second statement to be true, only when x=1 will the result be a prime. :(


Because \(\sqrt{x^2}=|x|\), not x. Try plugging in a negative numbers to check.

Why does \(\sqrt{x^2}=|x|\)?

The point here is that square root function cannot give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=|x|\)
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Re: M28-48  [#permalink]

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New post 09 Jan 2019, 01:14
Bunuel wrote:
If \(x\) is an integer, what is the value of \(x\)?


(1) \(|23x|\) is a prime number.

(2) \(2\sqrt{x^2}\) is a prime number.


#1 :
x can be +/-1 so insufficeint
#2:
again x can be +/-1 in sufficeint

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Re: M28-48  [#permalink]

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New post 02 Oct 2019, 06:35
Hi
Could anyone please explain here that when in second statement: 2 sqrt(x2) is said to be prime number, doesn't that mean we have to take value of x = +1
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New post 02 Oct 2019, 06:40
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Re: M28-48   [#permalink] 02 Oct 2019, 06:40
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