[tex] \sqrt{2x^2+8} =x-2[/tex]
Conditiile de existenta ale radicalului:
[tex]2x^2+8 \geq 0\rightarrow x \in R\\
x-2 \geq 0 \rightarrow x \geq 2[/tex]
x ∈ [2, ∞)
Ridicam la patrat in ambii mambri:
[tex]2x^2+8 =(x-2)^2\\ 2x^2+8=x^2-4x+4\\ x^2+4x+4=0\\ (x+2)^2=0\\x=-2[/tex]
Dar -2 ∉ [2, ∞) ==> x ∈ Ф
