1958 AHSME Problems/Problem 39

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 $\fbox{C}$ .

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.

Page

Toolbox

Search

1958 AHSME Problems/Problem 39

Problem

Solution

See Also