Difference between revisions of "1981 AHSME Problems/Problem 1"
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==Solution 1== | ==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> | |
==Solution 2== | ==Solution 2== |
Revision as of 13:56, 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
We can immediately get that , after we square
, we get
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.