Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.2 AMC 10/12 Find the answer

We may say concerning the solution of $|x|^2 + |x| - 6 =0$ that:
$\textbf{(A)}\ \text{there is only one root}$ $\textbf{(B)}\ \text{the sum of the roots is }{+1}$ $\textbf{(C)}\ \text{the sum of the roots is }{0}$ $\\ \textbf{(D)}\ \text{the product of the roots is }{+4}$ $\textbf{(E)}\ \text{the product of the roots is }{-6}$

This was a multiple-choice question, but the options didn't survive into the source we have. The answer given is C, and the solution below works it through.

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 $\fbox{C}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.