I have to solve $x^2 + 4x + 4 = 7|x+2|$.

I did this: $(x + 2)^2 = 7|x+2|$

And we know that $|w| = w \iff w ≥ 0$, so:

$|x+2|^2 = 7|x+2|$ because the $(x+2)^2$ is always $≥0$

Then, I divided this equation by $|x+2|$ (I think I can, because the $|x+2|^2 = |x+2||x+2|$

so I got $|x+2|=7$

It means that $x = 5$ or $x = -9$. But it's bad, because the valid result is $x=-9$ or $x=5$ or $x=-2$.

I know another approach that will give appropriate result, but why this one doesn't work?

  • 13
    $\begingroup$ You should've factored instead of divided, because you can't divide by a potential zero. That's how you lose solutions. $\endgroup$ Sep 18 at 16:32

4 Answers 4


The mistake turns out not to have anything to do with the absolute values directly. Dividing both sides by an expression is only valid when that expression is nonzero. So as soon as we divide by $|x+2|$, we have to consider a separate case where $|x+2|=0$; that recovers the solution $x=-2$.


Instead of division, you can use factorisation: \begin{align}|x+2|^2 = 7|x+2|&\iff |x+2|^2-7|x+2| = 0\\ &\iff|x+2|(|x+2|-7) = 0\\ &\iff |x+2| = 0\text{ or } |x+2|= 7 \end{align} Starting from $|x+2| = 7$, you can multiply it by $|x+2|$ to get $|x+2|^2 = 7|x+2|$, so $|x+2| = 7 \implies |x+2|^2=7|x+2|$, but the other direction is not true because you can only divide by $|x+2|$ if it is not equal to $0$. So, you need to consider two cases, when $|x+2| \neq 0$ in which case you are allowed to divide, and the second case $|x+2| = 0$, in which you are not, but it leads to solution $x = -2$ anyway.

Let's see another example: $x^2 - x = 0$. If you factor it, you get $x(x-1) = 0$, so $x = 0$ or $x = 1$ (you can get the same from quadratic formula). But, if you instead write $x^2 = x$ and then divide by $x$, you get $x = 1$ and lose the solution $x = 0$.


You can also break it up into cases.

Case $0: |x+2|=0\implies x=-2$


Case $1: x+2>0$:

$(x+2)^2=7(x+2)\implies x+2=7\implies x=5$

Case 2: $x+2<0$:

$(x+2)^2=-7(x+2)\implies x+2=-7\implies x=-9$

Then take the union.

  • 1
    $\begingroup$ You've considered $|x+2|=0$ three times. $\endgroup$
    – Ennar
    Sep 20 at 8:43

You can prevent a loss of solution here by taking everything into the LHS and then factoring out $|x+2|$. Dividing here is not wrong, you just have to ensure in your final answer that $-2$ is mentioned along with the others that you obtained.


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