Question

Rezolvati ecuația radical 2x patrat +8 (tot sub radical) =x-2

Answer 1

$\sqrt{2x^2+8} =x-2$

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:

$2x^2+8 =(x-2)^2\\ 2x^2+8=x^2-4x+4\\ x^2+4x+4=0\\ (x+2)^2=0\\x=-2$

Dar -2 ∉ [2, ∞)  ==>  x ∈ Ф