How do you solve and check for extraneous solutions in abs(x + 6) = 2x|x+6|=2x?

1 Answer
Stefan V.
Aug 2, 2015

x = 6x=6

Explanation:

Your absolute value equation looks like this

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

Right from the start, you can say that any negative value of xx will be an extraneous solution because the absolute value of a number can only be positive.

So, you need to check two cases for your equation

  • If (x+6)>0(x+6)>0, you have

|x+6| = x+6|x+6|=x+6

The equation becomes

x+6 = 2x => x = color(green)(6)x+6=2x⇒x=6

  • If (x+6)<0(x+6)<0, you have

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

The equation will be

-x-6 = 2x => 3x = -6 => x = (-6)/3 = color(red)(-2)−x−6=2x⇒3x=−6⇒x=−63=−2

This solution will be extraneous because it implies that the absolute values of 44 is negative, which is false.

|color(red)(-2) + 6| = 2 * color(red)((-2))|−2+6|=2⋅(−2)

|4| = -4 <=> 4!=-4|4|=−4⇔4≠−4

The first solution is valid, since you have

|color(green)(6) + 6| = 2 * color(green)(6)|6+6|=2⋅6

|12| = 12 <=> 12 = 12|12|=12⇔12=12

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