Question

resolva - 20 = - x - x ao quadrado​

Answer 1

Resposta:

Explicação passo-a-passo:

-20= -x-x²

x²+x-20=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+x-20=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=1~e~c=-20\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-20)=1-(-80)=81\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{81}}{2(1)}=\frac{-1-9}{2}=\frac{-10}{2}=-5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{81}}{2(1)}=\frac{-1+9}{2}=\frac{8}{2}=4\\\\S=\{-5,~4\}$

Answer 2

Resposta:

S={ -5 ; 4 }

Explicação passo-a-passo:

-20=-x-x²

x²+x-20=0

a=1

b=1

c=-20

∆=b²-4.a.c

∆=(1)²-4.(1).(-20)

∆=1+80

∆=81

x'=[-(1)+√81]/2

x'=[-1+9]/2

x'=8/2

x'=4

x"=[-(1)-√81]/2

x"=[-1-9]/2

x"=-10/2

x"=-5

Espero ter ajudado!