The set A = [x : x \(\in\) R, \(x^2\) = 16 and 2x = 6] equal

Solution :

\(x^2\) = 16 \(\implies\) x = \(\pm\)4

2x = 6 \(\implies\) x = 3

There is no value of x which satisfies both the above equations.

Thus, A = \(\phi\)

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