Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
Line 11: Line 11:
  
 
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>
 
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>
 +
 +
 +
 +
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 $|x|^2 + |x| - 6 =0$ 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 $\fbox{C}$


We can find all roots $x$ by setting $|x| = y$. This gives us the equation $y^2+y-6=0$, which has the solutions $y=-3, 2$. However, $|x|$ cannot equal $-3$, so the roots for $x$ are $2$ and $-2$. The sum of the two roots is $0$, making the answer $\fbox{C}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_39&oldid=92238"