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If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink ]
26 Dec 2012, 04:31

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If n and k are positive integers, is

\sqrt{n+k}>2\sqrt{n} (1) k > 3n

(2) n + k > 3n

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Re: If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink ]
26 Dec 2012, 04:36

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If n and k are positive integers, is (n+k)^1/2>2n^1/2? [#permalink ]
27 Jun 2014, 01:36

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!

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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
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 .

Hope it helps.

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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 .

Hope it helps.

I got it, thank you!!

_________________

Start to fall in love with GMAT <3

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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.

Thanks

.

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