Question

dá uma olhadinha quem manja de equação ae rsrsrs..!!​

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-4x+3=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-4~e~c=3\\\\\Delta=(b)^{2}-4(a)(c)=(-4)^{2}-4(1)(3)=16-(12)=4\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-4)-\sqrt{4}}{2(1)}=\frac{4-2}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-4)+\sqrt{4}}{2(1)}=\frac{4+2}{2}=\frac{6}{2}=3\\\\S=\{1,~3\}$

b)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+1x-30=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=1~e~c=-30\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-30)=1-(-120)=121\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{121}}{2(1)}=\frac{-1-11}{2}=\frac{-12}{2}=-6\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{121}}{2(1)}=\frac{-1+11}{2}=\frac{10}{2}=5\\\\S=\{-6,~5\}$

c)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+3x-10=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=3~e~c=-10\\\\\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(1)(-10)=9-(-40)=49\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{49}}{2(1)}=\frac{-3-7}{2}=\frac{-10}{2}=-5\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{49}}{2(1)}=\frac{-3+7}{2}=\frac{4}{2}=2\\\\S=\{-5,~2\}$