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

Revision as of 13:23, 28 June 2025

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

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

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

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

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 0	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

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 ==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>
+== See Also ==
+{{AHSME box|year=1981|num-b=0|num-a=2}}