Question

qual é a soluçao real da eguaçao x² - 8x + 12=0

Answer 1

Resposta:

Explicação passo-a-passo:

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}-8x+12=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=-8~e~c=12\\\\\Delta=(b)^{2}-4(a)(c)=(-8)^{2}-4(1)(12)=64-(48)=16\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-8)-\sqrt{16}}{2(1)}=\frac{8-4}{2}=\frac{4}{2}=2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-8)+\sqrt{16}}{2(1)}=\frac{8+4}{2}=\frac{12}{2}=6\\\\S=\{2,~6\}$