Question

Se considera functia f:R->R, f(x)=$x^{2}$-2x-8. determinati numerele intregi x pentru care f(x)<0.

Answer 1

x²-2x-8<0
Δ=4+32=36
x1=(2-6)/2=-2
x2=(2+6)/2=4
+++++-2----------4++++++
x∈(-2;4)
numerele intregi din acest interval sunt:x∈{-1;0;1;2;3}

Answer 2

x²-2x-8≤0
Δ=4+4×8=36
x1,2=2+sau-6/2
x1=-2
x2=4
deci x∈(-2;4)∩Z=-1,0,1,2,3