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1981 AHSME Problems/Problem 1

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Problem

If $\sqrt{x+2}=2$, then $(x+2)^2$ equals:

$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$


Solution 1

If we square both sides of the $\sqrt{x+2} = 2$, we will get $x+2 = 4$, if we square that again, we get $(x+2)^2 = \boxed{\textbf{(E) }16}$

Solution 2

We can immediately get that $x = 2$, after we square $(2+2)$, we get $\boxed{\textbf{(E) }16}$

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