Respuesta :

naǫ
[tex]6x-x^2-8 \ \textgreater \ 0 \\ -x^2+6x-8\ \textgreater \ 0 \\ -x^2+2x+4x-8\ \textgreater \ 0 \\ -x(x-2)+4(x-2)\ \textgreater \ 0 \\ (-x+4)(x-2)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ (-x+4)(x-2)=0 \\ -x+4=0 \ \lor \ x-2=0 \\ x=4 \ \lor \ x=2[/tex]

The coefficient of x is negative, so the parabola opens downwards. The values greater than 0 are between the zeros.
The solution set for the inequality is:
[tex]x \in (2,4)[/tex]

The answer is B.
-x²+6x-8>0           a=-1 , b=6    , c=-8
Δ = b² - 4.a.c 
Δ = 6² - 4 . -1 . -8 
Δ = 36 - 32
Δ = 4

x = (-b +- √Δ) / 2a

x' = (-6 + √4) / 2(-1)        x'' = (-6 - √4)/ 2(-1)
x' = -4 / -2                     x'' = -8 / -2
x' = 2                            x'' = 4


2>x>4   ou   (2,4)
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