1981 AHSME Problems/Problem 1

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

Square both sides of the equation to get get $x+2 = 4$. Then square both sides again to get $(x+2)^2 = \boxed{\textbf{(E) }16}$

Solution 2

Solve the equation to find that $x = 2$. By substituting, we find that $(x+2)^2=(2+2)^2=4^2=\boxed{\textbf{(E) }16}$

See Also

1981 AHSME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png