Question

4) Resolva as equações do 2º grau aplicando a fórmula de Bhaskara

a) X² - 3x +2 = 0

b) X² - 8x +12 = 0

Answer 1

Resposta:

Explicação passo a passo:

Resolva as equações do 2º grau aplicando a fórmula de Bhaskara

a) X² - 3x +2 = 0

a = 1; b = - 3; c = 2

/\= b^2 - 4ac

/\= (-3)^2 - 4.1.2

/\= 9 - 8

/\= 1

X = [- b +/- \/ /\] / 2a

X = [-(-3) +/- \/1]/2.1

X' = (3+1)/2= 4/2= 2

X" = (3-1)/2= 2/2= 1

R.:

S = {1;2}

_______________

b) X² - 8x +12 = 0

a = 1; b = - 8; c = 12

/\= b^2 - 4ac

/\= (-8)^2 - 4.1.12

/\= 64 - 48

/\= 16

X = [ - b +/- \/ /\] / 2a

x = [-(-8)+/- \/16]/2.1

X = (8+/- 4)/2

X ' = (8+4)/2 = 12/2= 6

X " = (8-4)/2 = 4/2= 2

R.:

S = {2, 6}

Answer 2

Resposta:

$x^2-3x+2=0\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-3\right)\pm \:1}{2\cdot \:1}\\\\\mathrm{Separe\:as\:solucoes}\\\\x_1=\frac{-\left(-3\right)+1}{2\cdot \:1},\:x_2=\frac{-\left(-3\right)-1}{2\cdot \:1}\\\\\mathrm{As\:solucoes\:para\:a\:equacao\:de\:segundo\:grau\:sao:\:}\\\\x=2\\\\\:x=1\\\\\\x^2-8x+12=0\\\\x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\cdot \:1\cdot \:12}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-\left(-8\right)\pm \:4}{2\cdot \:1}\\\\\mathrm{Separe\:as\:solucaes}\\\\x_1=\frac{-\left(-8\right)+4}{2\cdot \:1},\:x_2=\frac{-\left(-8\right)-4}{2\cdot \:1}\\\\\mathrm{As\:solucoes\:para\:a\:equacao\:de\:segundo\:grau\:sao:\:}\\\\x=6\\\\\:x=2$