Question

a) x² - 16x =0

b) x² + 9x + 14 = 0


ajudem ai essa atividade vale ponto '- ​

Answer 1

Resposta:

a) S={0, 16}

b) S={-7, -2}

Explicação passo a passo:

a) x²-16x=0

x(x-16)=0

1a solução:

x=0

2a solução:

x-16=0

x=16

S={0, 16}

Ou

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-16x-0=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-16~e~c=0\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-16)^{2}-4(1)(0)=256-(0)=256\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-16)-\sqrt{256}}{2(1)}=\frac{16-16}{2}=\frac{0}{2}=0\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-16)+\sqrt{256}}{2(1)}=\frac{16+16}{2}=\frac{32}{2}=16$

b) x²+9x+14=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}+9x+14=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=9~e~c=14\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(9)^{2}-4(1)(14)=81-(56)=25\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(9)-\sqrt{25}}{2(1)}=\frac{-9-5}{2}=\frac{-14}{2}=-7\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(9)+\sqrt{25}}{2(1)}=\frac{-9+5}{2}=\frac{-4}{2}=-2\\\\S=\{-7,~-2\}$