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4x^2-12x+8=0
ajutor
dau coroana


Răspuns :

[tex]4x^2-12x+8=0=\ \textgreater \ \\ 4(x^2-3x+2)=0=\ \textgreater \ \\ 4(x^2-x-2x+2)=0=\ \textgreater \ \\ 4(x-2)(x-1)=0 =\ \textgreater \ \\ (x-2)(x-1)=0=\ \textgreater \ \\ x-2=0 =\ \textgreater \ \\ x-1=0=\ \textgreater \ \\ x_1=2; \\ x_2=1 \\ S=[1;2] \\ Succes[/tex]