Difference between revisions of "1981 AHSME Problems/Problem 1"
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| + | ==Problem== | ||
If <math>\sqrt{x+2}=2</math>, then <math> (x+2)^2 </math> equals: | If <math>\sqrt{x+2}=2</math>, then <math> (x+2)^2 </math> equals: | ||
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<math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 </math> | <math> \textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16 </math> | ||
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| + | |||
| + | ==Solution 1== | ||
| + | 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> | ||
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| + | ==Solution 2== | ||
| + | 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|before=First question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:57, 29 June 2025
Contents
Problem
If 
, then 
equals:
Solution 1
Square both sides of the equation to get get 
. Then square both sides again to get 
Solution 2
Solve the equation to find that 
. By substituting, we find that 
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
