Question

quais sao as raizes da equaçao ?

urgente me ajuda por favor ​

Answer 1

Resposta:

Explicação passo-a-passo:

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-3x-4=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-3~e~c=-4\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4\\\\S=\{-1,~4\}$

Answer 2

Resposta:

letra D

Explicação passo-a-passo:

-3-4.1.(-4)

9+16

25

x= 3 +ou-5/2

x1= 8/2=4

x2=-2/2 =-1