Difference between revisions of "1978 AHSME Problems/Problem 1"
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<math>\frac{2}{x} = \boxed{\textbf{(B) }1}</math> | <math>\frac{2}{x} = \boxed{\textbf{(B) }1}</math> | ||
~awin | ~awin | ||
+ | |||
+ | ==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> |
Revision as of 21:16, 25 February 2020
Contents
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