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rezolvati in multimea nr complexe : (x^2-2x)^2 +x^2-2x=2

Răspuns :

Lucasela

(x^2-2x)^2 +(x^2-2x)=2

notam y= x^2-2x

=> y^2+y-2=0

Δ=1+8=9

y1=(-1+3)/2=1

y2=(-1-3)/2=-2

revenim,

x^2-2x=1=> x^2-2x-1=0

Δ=4+4=8

x1=(2+2√2)/2=1+√2

x2=(2-2√2)/2=1-√2

x^2-2x= -2

x^2-2x+2=0

Δ=4-8= -4

x3=(2+2i)/2=1+i

x4=(2-2i)/2=1-i


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