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.
Start to fall in love with GMAT <3