Question

8x2 – 8x = 0
Usando bhaskara como faço ?

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~8x^{2}-8x-0=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=8{;}~b=-8~e~c=0\\\\\Delta=(b)^{2}-4(a)(c)=(-8)^{2}-4(8)(0)=64-(0)=64\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-8)-\sqrt{64}}{2(8)}=\frac{8-8}{16}=\frac{0}{16}=0\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-8)+\sqrt{64}}{2(8)}=\frac{8+8}{16}=\frac{16}{16}=1\\\\S=\{0,~1\}$

Answer 2

Resposta:

∆=b²-4.a.c

∆=-8²-4.8.0

∆=64

x=-b±√∆/2.a

x=8±8/16

x'=8+8/16

x'=16/16

x'=1

x"=8-8/16

x"=0/16

x"=0