How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring? 
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong. 
 A: Hint: Note that the right-hand side must be $x+4\geq 0$ or the equality could not be true because the left-hand side is always positive. What can you conclude for $|x+4|$ in this case?
A: Hint: Observe that $x+4\ge0$  thus we can get rid of one absolute value and write $(x+4)|x-4|=(x+4)$ or $(x+4)(|x-4|-1)=0$. Squaring may create false roots.
A: Do squaring!
$$ (\vert x- 4 \vert \cdot \vert x + 4 \vert)^2 = (x + 4)^2 $$
$$ (\vert x- 4 \vert)^2 \cdot (\vert x + 4 \vert)^2 = (x + 4)^2 $$
$$ (x- 4)^2 \cdot ( x + 4)^2 = (x + 4)^2 $$
$$ (x- 4)^2 \cdot ( x + 4)^2 - (x + 4)^2=0 $$
$$ (x + 4)^2  \left[(x - 4)^2 - 1\right] =0 $$
$$ (x+4)^2=0, (x-4)^2 - 1 = 0$$
$$\implies x= -4, x = 4±\sqrt{1}$$
$$\implies x= -4, x = 5,x= 3$$

Njoy!
A: You can split the problem into 3 cases: $x\lt-4$, $-4\le x\le4$, and $4\lt x$.
For $x \lt -4$:


*

*There are no solutions for this case because $|x+4|*|x-4|$ is always positive, and $x+4$ will always be negative for $x<-4$.


For $-4\le x\le 4$:


*

*$|x-4| = 4-x$, and $|x+4| = x+4$, so solve $(4-x)*(x+4)=x+4$.

*$-x^2+16 = x+4$

*$0=x^2+x-12$

*$x = 3, -4$
For $4\lt x$:


*

*$|x-4| = x-4$, and $|x+4| = x+4$, so solve $(x-4)*(x+4)=x+4$.

*$x^2-16 = x+4$

*$0=x^2-x-20$

*$x = 5, -4$
All solutions: $x = -4, 3, 5$
A: Given
$$
|x+4||x-4|=x+4,
$$
and we want to find $x$, we observe that 
$$
|x+4||x-4|=|(x+4)(x-4)|=|x^2+4x-4x-16|=|x^2-16|.
$$
There are two cases:


*

*$x^2-16\geq 0$.  This happens when $x\geq 4$ or $x\leq -4$.  When $x^2-16\geq 0$, $|x^2-16|=x^2-16$ since both are nonnegative.  Therefore, you want to solve $x^2-16=x+4$.  Now, use the quadratic formula, you can get the solutions (but you'll need to check them against the inequalities above).

*$x^2-16<0$.  This happens when $-4<x<4$.  When $x^2-16<0$, then $|x^2-16|=-(x^2-16)=16-x^2$.  This is true since $x^2-16<0$ but the absolute value makes it positive by multiplying by $-1$.  Again, you need to use the quadratic formula to solve $16-x^2=x+4$ (and again, discard any values that you get that violate $-4<x<4$).
By other's observations on the positivity, you only need to consider the case where $x\geq -4$, but that only slightly simplifies this approach.
A: Note that if $x+4 > 0$ then $|x+4| = x+4>0$ and your equation reduces to $|x-4|=1$, which implies $x=5$ or $x=3$.
Alternatively, $x+4=0$, in which case $|x+4| = x+4 =0$, so the equation holds and $x=-4$.
Finally, if $x+4 < 0$ then we get $|x-4|=-1$ which has no real solutions.
Hence, the entire set of solutions is given by $-4, 3, 5$.

If you want to find the solutions by squaring, note that $|a|^2 = a^2$, therefore squaring both sides yields
$$
(x+4)^2(x-4)^2=(x+4)^2,
$$
which if $x \ne -4$ implies  $(x-4)^2=1$, yielding $x = 3$ or $x=5$. Alternatively, if $x=-4$, you get $=0$ so the equation holds and $x=-4$ is a solution.
in summary, you get the same solution set this way.

Note, however, that in general, squaring both sides of the equation may introduce additional solutions, which are not the solutions of the original equation. Consider, for example, the equation $x=1$, which trivially has a unique solution. Squaring both sides yields $x^2=1$, for which the original $x=1$ is a solution, but the second equation has an extra solution at $x=-1$, which does not solve the original equation.
As a result, one must use such techniques very carefully, checking all the solutions you find in the process to validate that they actually solve the original equation.
A: You can consider two cases:
If $\ x+4=0$, then $x=-4$. So
$$\vert{x+4}\vert(x-4)=0=x+4.$$
IF $\ x+4\neq 0$, then you can write
$$x-4=\frac{x+4}{x+4}=\mathrm{sign}(x+4)=\begin{cases} 1 &x+4\geq0\\ -1&x+4<0\end{cases}=\begin{cases} 1 &x\geq-4\\ -1&x<-4\end{cases}$$

from that the solutins are $x=3$ is $\ x=5$.

Or you can take squares like you said:
$$(x+4)^{2}(x-4)^{2}=(x+4)^{2}$$
and saying again if $x=-4$ both sides are equal to $0$. Otherwise you can cancel both factors $(x+4)^{2}$ (remember only when $x\neq -4$) to get:
$$ (x+4)^{2}(x-4)^{2}=(x+4)^{2} \overset{\displaystyle{x\neq-4}}{\Longrightarrow} (x-4)^2=1 \Longrightarrow x-4=\pm 1$$
so, $x-4=-1$ or $x-4=1$, so the 3 solutions are:
$$x_1=-4, x_2=3, x_3=5$$
