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

mathdad
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Solve the absolute value equation.

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

Seeking the first step.
 
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$x = 0$ is one solution.
Now divide both sides by $|x|$.

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.
 
|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.
 
RTCNTC said:
|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:

$$|a|=|b|$$

Then, this implies:

$$a=\pm b$$

And so, you have another case to evaluate:

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

$$2x=-4$$

$$x=-2$$

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:

$$x=\frac{-6+2}{2}=-2$$

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

$$x\in\{-2,0\}$$
 
What if the same question involves higher powers?

Example:

|x^2 - 2x|^3 = |x^2 + 6x|^3
 
RTCNTC said:
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 [math]|x|^3 = |x| \cdot x^2[/math]. It won't help you to get rid of the absolute value bars.

-Dan
 
topsquark said:
IF they have that as a given then you have to expand it out. But for the record [math]|x|^3 = |x| \cdot x^2[/math]. It won't help you to get rid of the absolute value bars.

-Dan
I will post a few more if needed.
 

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