How do you find the values that $x$ can take by squaring?

Question

$$|x+4| \cdot |x-4| = x+4$$

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.

Answer 1

Do squaring!

$$ (\vert x- 4 \vert \cdot \vert x + 4 \vert)^2 = (x + 4)^2 $$

$$ (\vert x- 4 \vert)^2 \cdot (\vert x + 4 \vert)^2 = (x + 4)^2 $$

$$ (x- 4)^2 \cdot ( x + 4)^2 = (x + 4)^2 $$

$$ (x- 4)^2 \cdot ( x + 4)^2 - (x + 4)^2=0 $$

$$ (x + 4)^2 \left[(x - 4)^2 - 1\right] =0 $$

$$ (x+4)^2=0, (x-4)^2 - 1 = 0$$

$$\implies x= -4, x = 4±\sqrt{1}$$

$$\implies x= -4, x = 5,x= 3$$

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Njoy!

Answer 2

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?

Answer 3

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.

Answer 4

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.

Answer 5

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: $$x_1=-4, x_2=3, x_3=5$$

Answer 6

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 $-4, 3, 5$.

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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.

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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.

Answer 7

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: $x = -4, 3, 5$