Difference between revisions of "1958 AHSME Problems/Problem 39"

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== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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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>
  
 
== See Also ==
 
== See Also ==

Revision as of 14:21, 22 February 2018

Problem

We may say concerning the solution of $|x|^2 + |x| - 6 =0$ that:

$\textbf{(A)}\ \text{there is only one root}\qquad  \textbf{(B)}\ \text{the sum of the roots is }{+1}\qquad  \textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\ \textbf{(D)}\ \text{the product of the roots is }{+4}\qquad  \textbf{(E)}\ \text{the product of the roots is }{-6}$

Solution

Note that for all roots $x$, $-x$ will also be a root. Therefore, the sum of all of the roots will be $0$, making the answer $\fbox{C}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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