1958 AMC 12 Problem 39

Question

We may say concerning the solution of

|x|^2+|x|-6=0

that:

there is only one root

the sum of the roots is

1

the sum of the roots is

0

the product of the roots is

4

the product of the roots is

-6

Set

u=|x|

, so

u\ge0

Factor the resulting quadratic in

u

, then translate the admissible value back to

x

Let

u=|x|\ge0.

Then

\begin{aligned} u^2+u-6&=0,\\ (u+3)(u-2)&=0. \end{aligned}

The nonnegative solution is

u=2,

so

|x|=2

and

x=\pm2.

Their sum is

0.

Therefore, the correct answer is C.

Problem 39 in other years