Difference between revisions of "1981 AHSME Problems/Problem 1"

Latest revision as of 13:57, 29 June 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 $x = 2$ . By substituting, we find that $(x+2)^2=(2+2)^2=4^2=\boxed{\textbf{(E) }16}$

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

@@ Line 6: / Line 6: @@
 ==Solution 1==
-If we square both sides of the <math>\sqrt{x+2} = 2</math>, we will get <math>x+2 = 4</math>, if we square that again, we get <math>(x+2)^2 = \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==
-We can immediately get that <math>x = 2</math>, after we square <math>(2+2)</math>, we get <math>\boxed{\textbf{(E) }16}</math>
+Solve the equation to find that <math>x = 2</math>. By substituting, we find that <math>(x+2)^2=(2+2)^2=4^2=\boxed{\textbf{(E) }16}</math>
 == See Also ==
-{{AHSME box|year=1981|num-b=0|num-a=2}}
+{{AHSME box|year=1981|before=First question|num-a=2}}
+{{MAA Notice}}