1981 AHSME Problems/Problem 1: Difference between revisions

Latest revision as of 14:52, 18 August 2025

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

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

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

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

1981 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AHSME Problems and Solutions

@@ Line 5: / Line 5: @@
-==Solution==
+==Solution 1==
-<math>\boxed{\textbf{(E) }16}</math>
+Square both sides of the equation to get get <math>x+2 = 4</math>.  Then square both sides again to get <math>(x+2)^2 = \boxed{\textbf{(E) }16}</math>
+==Solution 2==
+Solve the equation to find that <math>x = 2</math>. Then <math>(x+2)^2=(2+2)^2=4^2=\boxed{\textbf{(E) }16}</math>
+== See Also ==
+{{AHSME box|year=1981|before=First question|num-a=2}}
+{{MAA Notice}}