Quadratic equation with absolute values

Question

Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$

My attempt:

$|x^{2}+x-2|= x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$

$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$

$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$ $|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$

$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$

$x^{2}+x-2+x^{2}-x-2=2$

$\Rightarrow 2x^{2}-6=0$

$\Rightarrow x^{2}=3$

$\Rightarrow x_1=\sqrt{3}, x_2=-\sqrt{3}$

$2)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-1,2) \Rightarrow x\in[1,2)$

$x^{2}+x-2-x^{2}+x+2=2$

$\Rightarrow 2x=2$

$\Rightarrow x=1$

$3)$ $x\in(-2,1)$ and $x\in(-\infty,-1]\cup[2,\infty)$ $\Rightarrow x\in(-2,-1]$

$-x^{2}-x+2+x^{2}-x-2=2$

$\Rightarrow x=-1$

$4)$ $x\in(-2,1)$ and $x\in(-1,2)$ $\Rightarrow x\in(-1,1)$

$-x^{2}-x+2-x^{2}+x+2=2$

$\Rightarrow -2x^{2}+2=0$

$\Rightarrow -2x^{2}=-2$

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm{1}$ but that's not in the interval $(-1,1)$ so I can throw away that solution.

My question is how do I find the intersection in $1)$ $x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$ since $-\sqrt{3}$ and $\sqrt{3}$ should not be the solutions.

Answer 1

Your work is essentially correct but, as you noted, the case 1) is not well solved. In this case the range is

$x\in(-\infty,-2]\cup[1,\infty)$ and $x\in(-\infty,-1]\cup[2,\infty)$, that is : $$ \{(-\infty,-2]\cup[1,\infty)\}\cap\{(-\infty,-1]\cup[2,\infty)\}=(-\infty,-2]\cup [2,\infty)=D_1 $$ that is: $x\le -2 \mbox{ or } x\ge 2$

since $-2<-\sqrt{3}<2$ and $-2<\sqrt{3}<2$ , these two values are not in $D_1$ and are not roots of the equation.

Answer 2

It is a bit hard to follow. Perhaps we could write $$ \begin{align} f(x)&=x^2+x-2\\ g(x)&=x^2-x-2 \end{align} $$ and solve $|f(x)|+|g(x)|=2$ as you did, case by case: $$ \begin{align} +f(x)+g(x)&=2\implies x\in\{-\sqrt3,\sqrt3\}\\ +f(x)-g(x)&=2\implies x=1\\ -f(x)+g(x)&=2\implies x=-1\\ -f(x)-g(x)&=2\implies x\in\{-1,1\} \end{align} $$ And then since $$ \begin{align} f(x)&=0\implies x\in\{-2,1\}\\ g(x)&=0\implies x\in\{-1,2\} \end{align} $$ where $f,g$ are quadratic functions pointing in the positive $y$-direction, we get the following sign table $$ \begin{array}{|c|c|} \hline x&&-2&&-1&&1&&2&\\ \hline f(x)&+&0&-&-&-&0&+&+&+\\ \hline g(x)&+&+&+&0&-&-&-&0&+\\ \hline \end{array} $$ and then we can simply check the solutions up against this table directly. Since $\pm\sqrt3$ fall in between $-2,-1$ or $1,2$ where $f$ and $g$ have opposite signs, they cannot be valid solutions, since they are solutions to the first equation, where $f,g$ both have positive sign.

Note: Zero can be taken as having any sign you want since $+0=-0$, so all the other solutions are valid, which makes the solution set $x\in\{-1,1\}$.

Answer 3

The derivative of the function is

$$(2x+1)\text{sign}|x^2+x-2|+(2x-1)\text{sign}|x^2-x-2|.$$

Depending on the signs, this reduces to $\pm4x$ or $\pm2$ so that the only possible root occurs at $x=0$. The changes of sign can occur at $-2,-1,1,2$ so that the function is certainly monotonous in the intervals

$$]-\infty,-2,-1,0,1,2,\infty[.$$

The endpoints correspond to the function values

$$\infty,4,\color{green}2,4,\color{green}2,4,\infty$$ and you clearly see the two solutions.