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

Answer 1

#x = 6#

Your absolute value equation looks like this

#|x+6| = 2x#

Right from the start, you can say that any negative value of #x# 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

#|x+6| = x+6#

The equation becomes

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

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

The equation will be

#-x-6 = 2x => 3x = -6 => x = (-6)/3 = color(red)(-2)#

This solution will be extraneous because it implies that the absolute values of #4# is negative, which is false.

#|color(red)(-2) + 6| = 2 * color(red)((-2))#

#|4| = -4 <=> 4!=-4#

The first solution is valid, since you have

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

#|12| = 12 <=> 12 = 12#

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Answer 2

Savannah Anderson

To solve abs(x + 6) = 2x, first, split the equation into two cases: x + 6 = 2x and -(x + 6) = 2x. Solve each case separately. After solving, check each solution in the original equation to ensure it is valid and not an extraneous solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.