Question

Resolva as equações do 2° grau do tipo ax²+bx+c=0

A) x²-7x+6=0

B)-x²+2x+35=0

C)-3x²+7x-2=0

D)0,1x²-0,7x+1=0

Answer 1

Resposta:

a)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-7x+6=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-7~e~c=6\\\\\Delta=(b)^{2}-4(a)(c)=(-7)^{2}-4(1)(6)=49-(24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-7)-\sqrt{25}}{2(1)}=\frac{7-5}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-7)+\sqrt{25}}{2(1)}=\frac{7+5}{2}=\frac{12}{2}=6\\\\S=\{1,~6\}$

b)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-x^{2}+2x+35=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-1{;}~b=2~e~c=35\\\\\Delta=(b)^{2}-4(a)(c)=(2)^{2}-4(-1)(35)=4-(-140)=144\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(2)-\sqrt{144}}{2(-1)}=\frac{-2-12}{-2}=\frac{-14}{-2}=7\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(2)+\sqrt{144}}{2(-1)}=\frac{-2+12}{-2}=\frac{10}{-2}=-5\\\\S=\{7,~-5\}$

c)

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~-3x^{2}+7x-2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=-3{;}~b=7~e~c=-2\\\\\Delta=(b)^{2}-4(a)(c)=(7)^{2}-4(-3)(-2)=49-(24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(7)-\sqrt{25}}{2(-3)}=\frac{-7-5}{-6}=\frac{-12}{-6}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(7)+\sqrt{25}}{2(-3)}=\frac{-7+5}{-6}=\frac{-2\div2}{-6\div2}=\frac{1}{3}\\\\S=\{2,\frac{1}{3}\}$

d)

0,1x²-0,7x+1=0

Multiplicando tudo por 10

x²-7x+10=0

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