# find the solution set of the following equation with absolute value

WA online me results $x = -5$, failed to reach it

$$\left ( \left | x \right |+2 \right )\left ( \left| x-2 \right| -3\right)=x^{2}+3$$

• Fernando, your LaTeX did not work because you had a \right in the wrong place. Please confirm that this edit was your intention. You had it in front of a $-3$ instead of in front of the right parentheses. Oct 13, 2015 at 0:56
• Also, the \left and \right were unnecessary in this case. We usually use those when there are fractions, exponentiations, subscripts, and that sort of thing to make the bars and parentheses automatically size correctly. Side note: Not seeing an inequality here. Oct 13, 2015 at 1:07
• so is corrected Oct 13, 2015 at 1:40

You can break into cases.

Case 1: $x\leq 0$. In this case, $|x|=-x$ and $|x-2|=-x+2$

Case 2: $0\leq x\leq 2$. In this case, $|x|=x$ and $|x-2|=-x+2$

Case 3: $2\leq x$. In this case, $|x|=x$ and $|x-2|=x-2$

For each case, you can try to solve the equality making the necessary replacements, looking only for solutions that match the case you are in.

For example, in case $3$, you have

$(x+2)(x-2-3)=x^2+3$ which simplifies to

$x^2-5x+2x-10=x^2+3$

$3x=-13$

$x=\frac{-13}{3}$

However, since we were trying to work in the case that $2\leq x$, we have that $\frac{-13}{3}$ is not a valid solution, and so there are no solutions to our original equation with $x\geq 2$.

Continuing through the other cases, you will find the result.