Difference between revisions of "1955 AHSME Problems/Problem 12"
Awesomechoco (talk | contribs) (→Solution 1) |
Awesomechoco (talk | contribs) (→Solution 1) |
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the equation is false. | the equation is false. | ||
Therefore the answer is <math>\fbox{{\bf(D)} x=1}</math> | Therefore the answer is <math>\fbox{{\bf(D)} x=1}</math> | ||
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+ | Solution by awesomechoco | ||
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+ | ==Solution 2== | ||
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+ | ==See Also== | ||
+ | == See Also == | ||
+ | {{AHSME box|year=1955|num-b=11|num-a=13}} | ||
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+ | {{MAA Notice}} |
Revision as of 00:37, 10 July 2018
Contents
[hide]Problem
The solution of is:
Solution 1
First, square both sides. This gives us
Then, adding to both sides gives us
After that, adding to both sides will give us
Next, we divide both sides by 2 which gives us
Finally, solving the equation, we get
Plugging 1 and 2 into the original equation, , we see that when
the equation is true. On the other hand, we note that when
the equation is false. Therefore the answer is
Solution by awesomechoco
Solution 2
See Also
See Also
1955 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.