Solve this inequality? #(x+1)^2 - abs(x-2) >= 0#

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Answer 1 · 2018-02-20T20:43:11

#x > 1/2(sqrt13-3)#

#(x+1)^2 - abs(x-2) >= 0# or

#(x+1)^2 ge abs(x-2)# and squaring both sides

#(x+1)^4 ge (x-2)^2# or

#(x+1)^4 - (x-2)^2 ge 0# or

#((x+1)^2+x-2)((x+1)^2-x+2) ge 0# or

#(x^2+3x-1)(x^2+x+3) ge 0#

now we have that #x^2+x+3 > 0 forall x# then the condition reduces to

#x^2+3x-1 ge 0# or

#{x < -1/2(3+sqrt13)} uu {x > 1/2(sqrt13-3)}#

and the feasible solution is

#x > 1/2(sqrt13-3)# verified by substitution.

NOTE

The squaring operation introduces extraneous additional solutions.