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sofyamorozova14
@sofyamorozova14
July 2022
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найдите множество корней уравнения x^2-(p+2)x+2p=0
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MizoriesKun
Х²-(р+2)*х+2р=0
D=(p+2)²-4*2p=p²+2p+4-8p =p²-4p+4 =(p-2)² √D=p-2
x₁ = (p+2)+(p-2)=p+2+p-2=2p
x₂ = (p+2)-(p-2)=p+2-p+2 =4
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Answers & Comments
D=(p+2)²-4*2p=p²+2p+4-8p =p²-4p+4 =(p-2)² √D=p-2
x₁ = (p+2)+(p-2)=p+2+p-2=2p
x₂ = (p+2)-(p-2)=p+2-p+2 =4