\(\)If d is a positive integer, is \(\sqrt{d}\) an integer?
(1) d is the square of an integer.
(2) \(\sqrt{d}\) is the square of an integer.
Sol: Given d is a postive integer we need to check whether \(\sqrt{d}\) = I where I is an integer
st1: Given d=a^2 where " a" is some postive integer then \(\sqrt{d}\) =a which is an integer So St 1 is sufficent
St2: given \(\sqrt{d}\) = b^2 where b is some integer so \(\sqrt{d}\) is an integer.
Ans is D.
Note: • 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, \(\sqrt{25}\) , NOT +5 or -5. In contrast, the equation x^2=25 has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.
• Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).
Courtesy GMAT CLUB Math Book
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”