Question

alguem sabe resolver ?f(x)=x²-2x-3

Answer 1

Resposta:

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

Answer 2

a= 1
b= -2
c= -3
b2 - 4ac
4 - 4 (-3) (1)
16
x= -b +- raiz de delta
2a
2+-4
——-
2

2+4
——- = 3
2


2-4
—— = -1
2