Polynomial equations with absolute values

Question

How can I solve: $x^2 + 2|x| - 3 = 0$ ?

My attempt:

$|x| = \frac{3 - x^2}{2}$

$x = \pm \frac{3 - x^2}{2}$

$x^2 \pm 2x - 3 = 0$

The solutions to this 2nd degree polynomial is

$x_1 = -3$

$x_2 = 1$

$x_3 = -1$

$x_4 = 3$

However, only $-1$ and $1$ are solutions. Why?

Solve it as Bernard says and get two possible values of $|x|$ as $1$ and $-3$ . Then $|x|$ cannot be negative hence $|x|=1$ implying $x=\pm 1$ — user118494, Sep 03 '15 at 16:08

score 1 · Answer 1 · answered Sep 03 '15 at 16:02

1

From $|x|=\frac{3-x^2}{2}$, you have to have $\frac{3-x^2}{2}\ge 0$, i.e. $-\sqrt 3\le x\le \sqrt 3$.

answered Sep 03 '15 at 16:02

mathlove

score 1 · Answer 2 · answered Sep 03 '15 at 16:03

1

That is very complicated.

Simple hint:

$x^2=\lvert x\rvert^2$. You get a quadratic equation with unknown $\lvert x\rvert$.

answered Sep 03 '15 at 16:03

Bernard

score 0 · Answer 3 · answered May 30 '17 at 23:36

Well, for starters, I'd pretend that the absolute value sign wasn't there and solve it normally. We get $x^2$+2x+3 = (x+3)(x-1).

Since x cannot be negative in your actual equation, -3 can't possibly be a solution. Bernard's hint paves the way to solve the problem.

3 Answers3