Difference between revisions of "1958 AHSME Problems/Problem 39"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | |
| + | Note that for all roots <math>x</math>, <math>-x</math> will also be a root. Therefore, the sum of all of the roots will be <math>0</math>, making the answer <math>\fbox{C}</math> | ||
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| + | We can find all roots <math>x</math> by setting <math>|x| = y</math>. This gives us the equation <math>y^2+y-6=0</math>, which has the solutions <math>y=-3, 2</math>. However, <math>|x|</math> cannot equal <math>-3</math>, so the roots for <math>x</math> are <math>2</math> and <math>-2</math>. The sum of the two roots is <math>0</math>, making the answer <math>\fbox{C}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 13:24, 22 February 2018
Problem
We may say concerning the solution of 
that:
Solution
Note that for all roots 
, 
will also be a root. Therefore, the sum of all of the roots will be 
, making the answer 
We can find all roots by setting 
. This gives us the equation 
, which has the solutions 
. However, 
cannot equal 
, so the roots for are 
and 
. The sum of the two roots is , making the answer .
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
