What are the real solutions to the equation |x|^2 + 2|x|-? 3 = 0 ?

F. ±1
G. ±3
H. 1 and 3
J. ?1 and ?3
K. ±1 and ±3

I tried factoring it and got 1 and -3, but that doesn't match any of the choices. How do I find the correct answer?

asked 22 Jan '16, 16:24

Sam_PrepScholar's gravatar image

Sam_PrepScholar
1.4k172174182


You're right that the equation factors out to (x-1)(x+3) if you ignore the absolute value signs. All you have to do is take one more step and plug the solutions back in.

If you put -3 into the original equation, it doesn't end up equalling zero because the absolute value signs turn the -3 into a regular positive 3.

3^2 + 2(3) -3 = 0
9 + 6 - 3 = 0
12 = 0 <------- NOPE!

Both 3 and -3 can be ruled out as solutions. Now, we can try plugging in 1. Even with the absolute value signs, 1 works as a solution.

1^2 + 2(1) -3 = 0
1 + 2 - 3 = 0
0 = 0 <------- YEP!

This means that -1 also works, because -1 and 1 have the same absolute value. The answer to the question is choice F.

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answered 22 Jan '16, 16:31

Sam_PrepScholar's gravatar image

Sam_PrepScholar
1.4k172174182

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