Question

(x-4)^2>sau=5 inecuatie

Answer 1

(x-4)²-5≥0
x²-8x+16-5≥0
x²-8x+11≥0
egalam cu 0.
a=1
b=-8
c=11
Δ=64-44
Δ=20
x1=$\frac{8+2 \sqrt{5} }{2}$=$\frac{2(4+ \sqrt{5)} }{2}$=4+√5.
x2=4-√5

facem tabel de semne

x             | -inf            4-√5         4+√5       +inf
x²-8x+11 |+++++++++0-------------0++++++++

solutie finala:  x∈(-inf; 4-√5] reunit cu [4+√5; +inf)