Question

X²+6x+8=0 e X²-10x-11=0 me ajudem e so esses dois.

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+6x+8=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=6~e~c=8\\\\\Delta=(b)^{2}-4(a)(c)=(6)^{2}-4(1)(8)=36-(32)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(6)-\sqrt{4}}{2(1)}=\frac{-6-2}{2}=\frac{-8}{2}=-4\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(6)+\sqrt{4}}{2(1)}=\frac{-6+2}{2}=\frac{-4}{2}=-2\\\\S=\{-4,~-2\}$

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-10x-11=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-10~e~c=-11\\\\\Delta=(b)^{2}-4(a)(c)=(-10)^{2}-4(1)(-11)=100-(-44)=144\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-10)-\sqrt{144}}{2(1)}=\frac{10-12}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-10)+\sqrt{144}}{2(1)}=\frac{10+12}{2}=\frac{22}{2}=11\\\\S=\{-1,~11\}$