Rozwiąż równanie (2x^2+3x+1)^2 - (4x^2 -5x +2)^2=0
ze wzoru a²-b²=(a+b)(a-b)
(2x²+3x+1+4x²-5x+2)(2x²+3x+1-4x²+5x-2)=0
(6x²-2x+3)(-2x²+8x-1)=0
Δ1=4-4*6*3=4-72=-68<0
brak pierwiastkow w I nawiasie
Δ=64-8=56
√Δ=2√14
x=(-8-2√14)/(-4)=2+1/2√14 v x=2-1/2√14
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
ze wzoru a²-b²=(a+b)(a-b)
(2x²+3x+1+4x²-5x+2)(2x²+3x+1-4x²+5x-2)=0
(6x²-2x+3)(-2x²+8x-1)=0
Δ1=4-4*6*3=4-72=-68<0
brak pierwiastkow w I nawiasie
Δ=64-8=56
√Δ=2√14
x=(-8-2√14)/(-4)=2+1/2√14 v x=2-1/2√14