Question

sabendo a função f(x) = - x² + 9x - 8,determine os valores reais de x para que se tenha:-

Answer 1

Resposta:

Explicação passo a passo:

f(x) = -x²+9x-8

a)

f(x) = 10

-x²+9x-8=10

-x²+9x-8-10=0

-x²+9x-18=0  ×(-1)

x²-9x+18=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-9x+18=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-9~e~c=18\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-9)^{2}-4(1)(18)=81-(72)=9\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-9)-\sqrt{9}}{2(1)}=\frac{9-3}{2}=\frac{6}{2}=3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-9)+\sqrt{9}}{2(1)}=\frac{9+3}{2}=\frac{12}{2}=6\\\\S=\{3,~6\}$

b)

f(x) = -15/4

-x²+9x-8 = -15/4 ×(-4)

4x²-36x+32=15

4x²-36x+32-15=0

4x²-36x+17=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~4x^{2}-36x+17=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=4{;}~b=-36~e~c=17\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-36)^{2}-4(4)(17)=1296-(272)=1024\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-36)-\sqrt{1024}}{2(4)}=\frac{36-32}{8}=\frac{4\div4}{8\div4}=\frac{1}{2}$$\displaystyle x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-36)+\sqrt{1024}}{2(4)}=\frac{36+32}{8}=\frac{68\div4}{8\div4}=\frac{17}{2}\\\\\\S=\{\frac{1}{2},~\frac{17}{2}\}$