If *x* is a perfect square (i.e., the square of some integer), then the next perfect square larger than *x* is:

*x*+ 1*x*^{2}+ 1*x*^{2}+ 2*x*+ 1*x*^{2}+*x**x*+ 2√*x*+ 1

Let *x* = *n*^{2}. Then √*x* = *n*. The next perfect square after *x* will be the square of *n* + 1. So we want (*n* + 1)^{2} in terms of *x*.

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*n*+ 1)^{2}=*n*^{2}+ 2*n*+ 1 =*x*+ 2√*x*+ 1. The answer is (e).Check: Let *x* = 49; then *n* = 7. So *n* + 1 = 8, and (*n* + 1)^{2} = 64.

Is 64 the next perfect square after 49? Yes.

Does *x* + 2√*x* + 1 = 64? Yes, 49 + 2 · 7 + 1 = 49 + 14 + 1 = 64.