Solve for #x#: #absx = abs(x-2) - abs(x+4)#?

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Answer 1 · 2017-11-03T01:13:35

See below.

The solutions for

#absx = abs(x-2) - abs(x+4)#

are included in the set of solutions for

#x^2=(x-2)^2+(x+4)^2-2 abs(x-2)abs(x+4)# or

#4(x-2)^2(x+4)^4=(x^2+4x+20)^2#

after simplifications

#(x-6) (2 + x) (6 + x) (2 + 3 x) = 0#

with the set of real solutions

#x = {-6,-2,-2/3,6}#

Now checking for feasibility we conclude that the feasible solutions are

#x = {-6, -2}#