1978 AHSME Problems/Problem 1: Difference between revisions
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==Solution 3== | ==Solution 3== | ||
Directly factoring, we get <math>(1-\frac{2}{x})^2 = 0</math>. Thus <math>\frac{2}{x}</math> must equal <math>\boxed{\textbf{(B) }1}</math> | Directly factoring, we get <math>(1-\frac{2}{x})^2 = 0</math>. Thus <math>\frac{2}{x}</math> must equal <math>\boxed{\textbf{(B) }1}</math> | ||
==See Also== | |||
{{AHSME box|year=1978|num-b=0|num-a=2}} | |||
{{MAA Notice}} | |||
Latest revision as of 10:59, 13 February 2021
Problem 1
If 
, then 
equals
Solution 1
By guessing and checking, 2 works.
~awin
Solution 2
Multiplying each side by 
, we get 
. Factoring, we get 
. Therefore, 
.
~awin
Solution 3
Directly factoring, we get 
. Thus must equal 
See Also
| 1978 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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