\(\)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

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