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