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Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]
27 Jun 2014, 01:53

Expert's post

LucyDang wrote:

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

Hi Bunuel,

I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive.

What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 --> \(\sqrt{x}\) = 3 or -3 Same thought or n!

Please help me to clarify, thank you so much!

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{9}=3\), NOT +3 or -3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and -3. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]
27 Jun 2014, 04:06

Bunuel wrote:

LucyDang wrote:

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

Hi Bunuel,

I don't understand why n & k are positive then two parts of the inequility \(\sqrt{n+k}>2\sqrt{n}\) are positive.

What I understand is that: n, k>0 => n+k>0 => \(\sqrt{n+k}\) might be positive or negative. e.g: x=9>0 --> \(\sqrt{x}\) = 3 or -3 Same thought or n!

Please help me to clarify, thank you so much!

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{9}=3\), NOT +3 or -3. In contrast, the equation \(x^2=9\) has TWO solutions, +3 and -3. Even roots have only non-negative value on the GMAT.

Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]
19 Aug 2014, 06:01

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n.

Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink]
19 Aug 2014, 07:06

1

This post received KUDOS

Expert's post

tobiasfr wrote:

Bunuel wrote:

If n and k are positive integers, is \(\sqrt{n+k}>2\sqrt{n}\)?

Both parts of the inequality are positive, thus we can square it, to get "is \(n+k>4n\)?" --> is \(k>3n\)?

(1) k > 3n. Sufficient.

(2) n + k > 3n --> \(k>2n\). Not sufficient.

Answer: A.

How do you get to the conclusion that \(\sqrt{n+k}>2\sqrt{n}\)? Even if k > 3n it can be either n+k > 4n or n+k < 4n, since we don't know more about n.

Thanks .

Not sure how you get the above...

Anyway, the question asks whether \(\sqrt{n+k}>2\sqrt{n}\)? After algebraic manipulations shown in my solution the question becomes: is \(k>3n\)? The first statement answers this question, which makes it sufficient. _________________

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