Question

Aplicando a fórmula de Bhaskara, resolva as seguintes equações do 2º grau.

A) 2x² – 6x + 4 = 0

B) 9x² – 12x + 4 = 0

C) x² - 3x - 4 = 0​

Answer 1

Resposta:

Explicação passo-a-passo:

a) 2x² – 6x + 4 = 0 (÷2)

x²-3x+2=0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-3x+2=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-3~e~c=2\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(2)=9-(8)=1\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{1}}{2(1)}=\frac{3-1}{2}=\frac{2}{2}=1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{1}}{2(1)}=\frac{3+1}{2}=\frac{4}{2}=2\\\\S=\{1,~2\}$

b) 9x² – 12x + 4 = 0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~9x^{2}-12x+4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=9{;}~b=-12~e~c=4\\\\\Delta=(b)^{2}-4(a)(c)=(-12)^{2}-4(9)(4)=144-(144)=0\\\\x^{'}=x^{''}=\frac{-(b)\pm\sqrt{\Delta}}{2(a)}=\frac{-(-12)\pm\sqrt{0}}{2(9)}=\frac{12}{18}=\frac{12\div6}{18\div6}=\frac{2}{3}\\\\S=\{\frac{2}{3},~\frac{2}{3}\}$

c) x² - 3x - 4 = 0​

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-3x-4=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-3~e~c=-4\\\\\Delta=(b)^{2}-4(a)(c)=(-3)^{2}-4(1)(-4)=9-(-16)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-3)-\sqrt{25}}{2(1)}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-3)+\sqrt{25}}{2(1)}=\frac{3+5}{2}=\frac{8}{2}=4\\\\S=\{-1,~4\}$