(x²-3x-4)(x²-7x+12)≤0
1) x²-3x-4=0
x₁=-1; x₂=4 (a-b+c=0 [1+3-4=0], x₁=-1; x₂=-c/a=4/1=4)
2) x²-7x+12=0
D=(-7)²-4*12=49-48=1
x₁=(7+1)/2=8/2=4
x₂=(7-1)/2=6/2=3
(x+1)(x-4)(x-3)(x-4)≤0
(x+1)(x-3)(x-4)²≤0
(x-4)²≥0 при x∈R
(x+1)(x-3)≤0
+ - +
..
-3 1
x∈[-3;1]
ответ: x∈[-3;1]
