" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
x₁=-3-5/2=-4
x₂=-3+5/2=1
f(x)=(x+4)(x-1)
f>0 <=> x∈(-nieskonczonosc, -4) suma (1, +nieskonczonosc)
f<0 <=> x∈(-4, 1)
a= -1 b=3 c=4
Δ=b²-4ac
Δ=3²-4*(-1)*4
Δ=9+16
Δ=25
√Δ=5
x₁=-b-√Δ/2=-3-5/2=-4
x₂=b+√Δ/2=-3+5/2=1
x₁ i x₂ to miejsca zerowe
f(X)=(x-x₁)(x-x₂)
podstawiamy za x₁ i x₂
f(x)=(x+4)(x-1) postać iloczynowa
f>0 <=> x∈(-∞ -4) suma (1, +∞)
f<0 <=> x∈(-4, 1)