Solve Absolute Value Equation |x^2 - 2x| = |x^2 + 6x|

[mathdad]

Solve the absolute value equation.

|x^2 - 2x| = |x^2 + 6x|

Seeking the first step.

 

[S.G. Janssens]

$x = 0$ is one solution.
Now divide both sides by $|x|$.

Spoiler

Next, write the resulting equation as $|x - 2| = |x - (-6)|$.
Then you seek those $x$ which have equal distance to the points $2$ and $-6$ on the real line.

 

[mathdad]

|x^2 - 2x| = |x^2 + 6x|

|x(x - 2)| = |x(x + 6)|

|x||x - 2| = |x||x+6|

[|x||x - 2|]/|x| = [|x||x+6|]/|x|

|x-2| = |x+6|

x - 2 = x + 6

The only solution is x = 0.

 

[MarkFL]

|x^2 - 2x| = |x^2 + 6x|

|x(x - 2)| = |x(x + 6)|

|x||x - 2| = |x||x+6|

[|x||x - 2|]/|x| = [|x||x+6|]/|x|

|x-2| = |x+6|

x - 2 = x + 6

The only solution is x = 0.

You can't just remove the absolute value as if they have no meaning. If we have:

\(\displaystyle |a|=|b|\)

Then, this implies:

\(\displaystyle a=\pm b\)

And so, you have another case to evaluate:

\(\displaystyle x-2=-(x+6)\)

\(\displaystyle 2x=-4\)

\(\displaystyle x=-2\)

Let's examine the hint given by Janssens:

"Then you seek those \(x\) which have equal distance to the points 2 and −6 on the real line."

This would be the mid-point or mean of the two stated values:

\(\displaystyle x=\frac{-6+2}{2}=-2\)

And so, we conclude that the original problem has two solutions, given by:

\(\displaystyle x\in\{-2,0\}\)

 

[mathdad]

What if the same question involves higher powers?

Example:

|x^2 - 2x|^3 = |x^2 + 6x|^3

 

[topsquark]

What if the same question involves higher powers?

Example:

|x^2 - 2x|^3 = |x^2 + 6x|^3

IF they have that as a given then you have to expand it out. But for the record \(\displaystyle |x|^3 = |x| \cdot x^2\). It won't help you to get rid of the absolute value bars.

-Dan

 

[mathdad]

IF they have that as a given then you have to expand it out. But for the record \(\displaystyle |x|^3 = |x| \cdot x^2\). It won't help you to get rid of the absolute value bars.

-Dan

I will post a few more if needed.