[tex](x+1)^2 \leq 9 \\ (x+1)^2-9 \leq 0 \\ x^2+2x+1-9 \leq 0 \\ x^2+2x-8 \leq 0[/tex]
Discriminantul este
[tex]D=4+32=36 \\ x_1= \frac{-2+ \sqrt{D} }{2}=2 \\ x_2= \frac{-2- \sqrt{D} }{2}=-4 [/tex]
Ne cere sa fie mai mic sau egal cu 0 ⇒ x∈[-4,2]
Dar x∈Z
Deci x∈{-4,-3,-2,-1,0,1,2}.
Pentru clasa a 6a:
[tex](x+1)^2 \leq 9 \\ Extragem\ radical \\ |x+1| \leq 3 \\ x+1 \leq 3\ sau\ x+1 \geq -3 \\ x \leq 2\ sau\ x \geq -4 [/tex]
Dar x∈Z ⇒ x∈{-4,-3,-2,-1,0,1,2}
