Rozwiąż: (x-1)^2=2x^2+1 ; x^2+4x+1=0
(x-1)^2=2x^2+1
x^2-2x+1 = 2x^2+1
x^2-2x-2x^2=0
-x^2-2x = 0 /:(-1)
x^2+2x = 0
x(x+2) = 0
x=0 lub x = -2
x^2+4x+1=0
Δ = 16 - 4 = 12
√Δ = √12 = √(4*3) = 2√3
x1 = (-4+2√3) / 2 = -2+√3
x2 = (-4-2√3) / 2 = -2-√3
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
(x-1)^2=2x^2+1
x^2-2x+1 = 2x^2+1
x^2-2x-2x^2=0
-x^2-2x = 0 /:(-1)
x^2+2x = 0
x(x+2) = 0
x=0 lub x = -2
x^2+4x+1=0
Δ = 16 - 4 = 12
√Δ = √12 = √(4*3) = 2√3
x1 = (-4+2√3) / 2 = -2+√3
x2 = (-4-2√3) / 2 = -2-√3