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 $\fbox{C}$.

See Also

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