f(x) < 0
x²-9x+8 < 0
Δ=(-9)²-4·1·8=81-32=49 , √Δ=√49=7
x1=(9-7)/2
x1=1
x2=(9+7)/2
x2=8
x∈(1,8)
[tex]f(x) = x^{2}-9x + 8\\\\x^{2}-9x + 8 < 0\\\\a = 1, \ b = -9, \ c = 8\\\\M. \ zerowe:\\\\\Delta = b^{2}-4ac = (-9)^{2} - 4\cdot1\cdot8 = 81 - 32 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-9) -7}{2\cdot1} = \frac{2}{2} = 1\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-9)+7}{2}=\frac{16}{2} = 8\\\\a > 0, \ ramiona \ paraboli \ skierowane \ do \ gory\\\\x \in (1;8)[/tex]
