Difference between revisions of "1958 AHSME Problems/Problem 39"
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<math> \textbf{(A)}\ \text{there is only one root}\qquad | <math> \textbf{(A)}\ \text{there is only one root}\qquad | ||
| − | \textbf{(B)}\ \text{the sum of the roots is }{ | + | \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad |
\textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ | \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ | ||
| − | \textbf{(D)}\ \text{the product of the roots is }{ | + | \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad |
| − | \textbf{(E)}\ \text{the product of the roots is }{ | + | \textbf{(E)}\ \text{the product of the roots is }{-6}</math> |
== 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 | |
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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. 
