Question

quais são as raízes da equação? (preciso das respostas completas)​

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$x^{2}-x-20=0\\\Delta=(b)^{2}-4(a)(c)=(-1)^{2}-4(1)(-20)=1-(-80)=81\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-1)-\sqrt{81}}{2(1)}=\frac{1-9}{2}=\frac{-8}{2}=-4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-1)+\sqrt{81}}{2(1)}=\frac{1+9}{2}=\frac{10}{2}=5\\S=\{-4,5\}$

b)

$x^{2}-3x-4=0\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4\\S=\{-1,4\}$

c)

$x^{2}-14x+48=0\\\Delta=(b)^{2}-4(a)(c)=(-14)^{2}-4(1)(48)=196-(192)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-14)-\sqrt{4}}{2(1)}=\frac{14-2}{2}=\frac{12}{2}=6\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-14)+\sqrt{4}}{2(1)}=\frac{14+2}{2}=\frac{16}{2}=8\\S=\{6,8\}$