Difference between revisions of "1953 AHSME Problems/Problem 4"
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The solutions are then <math>0, -4, 4</math>. So the answer is <math>\boxed{\text{C}}</math> | The solutions are then <math>0, -4, 4</math>. So the answer is <math>\boxed{\text{C}}</math> | ||
− | + | ==See Also== | |
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+ | {{AHSME 50p box|year=1953|num-b=3|num-a=5}} | ||
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+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:18, 1 April 2017
The roots of are:
Solution
We can divide both sides by to get the equation:
We factor the middle part into .
The equation now becomes
The solutions are then . So the answer is
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.