Question

Rezolvați inecuația:
(x²-3)(x²-3x-4)≥0

Answer 1

(x^2-3)(x^2-3x-4)>=0 <=> (x^2-3)(x^2-3x-4)=0 sau (x^2-3)(x^2-3x-4)>0.

(x^2-3)(x^2-3x-4)=0 <=> (x-rad(3))(x+rad(3))(x+1)(x-4)=0 <=>  x apartine {-rad(3),rad(3),-1,4}.

(x^2-3)(x^2-3x-4)>0 <=> (x^2-3<0 si x^2-3x-4<0) sau (x^2-3>0 si x^2-3x-4>0)

x^2-3<0 si x^2-3x-4<0 <=> x apartine (-rad(3), rad(3)) si x apartine (-1,4) <=> x apartine (-1,rad(3)).

x^2-3>0 si x^2-3x-4>0 <=> x apartine (-inf,-rad(3))U(rad(3),inf) si x apartine (-inf,-1)U(4,inf) <=> x apartine (-inf,-rad(3))U(4,inf).


In concluzie, multimea solutiilor inecuatiei din enunt este S=(-inf,-rad(3)]U[-1,rad(3)]U[4,inf).