How do you solve #|x|=-5x+24# and find any extraneous solutions?

1 Answer
Oct 31, 2017

Answer:

The piecewise definition of #|x| = {(x;x>=0),(-x;x<0):}# allows one to separate the equation into two equations, one with x and the other with -x and then solve both equations.
Check.

Explanation:

Separate #|x|=-5x+24# into two equations:

#x = -5x+24; x >=0# and #-x = -5x+24; x < 0#

Add 5x to both sides of both equations:

#6x = 24; x>=0# and #4x = 24;x<0#

Divide the first equation by 6 and the second equation by 4:

#x = 4; x>=0# and #x = 6;x<0#

We must discard the #x = 6# solution, because it is not within the domain restriction.

If we had not included the domain restrictions, we would have discovered the extraneous root by performing the check.

Check:

#|4|=-5(4)+24# and #|6| = -5(6)+24#

#|4|=4# and #|6| != -6#

Only the #x = 4# solution checks, therefore, we must discard the #x = 6# solution.