Respuesta :

fieryanswererft

Answer:

  • two solutions

x = 1, 3

Explanation:

[tex]\sf f(x)=x^2-4x+3[/tex]

[tex]\rightarrow \sf x^2-4x+3=0[/tex]

[tex]\rightarrow \sf x^2-3x-x+3=0[/tex]

[tex]\rightarrow \sf x(x-3)-1(x-3)=0[/tex]

[tex]\rightarrow \sf (x-1)(x-3)=0[/tex]

[tex]\rightarrow \sf x-1=0, \ x-3=0[/tex]

[tex]\rightarrow \sf x=1, \ x=3[/tex]

Thus, there are two solutions and the values are 1 and 3

Let's see

[tex]\\ \rm\rightarrowtail f(x)=0[/tex]

[tex]\\ \rm\rightarrowtail x^2-4x+3=0[/tex]

[tex]\\ \rm\rightarrowtail x^2-3x-x+3=0[/tex]

[tex]\\ \rm\rightarrowtail x(x-3)-1(x-3)=0[/tex]

[tex]\\ \rm\rightarrowtail (x-1)(x-3)=0[/tex]

[tex]\\ \rm\rightarrowtail x=1,3[/tex]

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