Question

dada a função quadrática f(x)=x²-7x+12,determine os pontos das raízes

Answer 1

$x^2-7x+12=0$

$\triangle=b^2-4.a.c=(-7)^2-4.1.12=49-48=1$

$x_1=\frac{-b+\sqrt{\triangle} }{2a}=\frac{7+\sqrt{1} }{2.1}=\frac{7+1}{2}=\frac{8}{2}=4$

$x_2=\frac{-b-\sqrt{\triangle} }{2a}=\frac{7-\sqrt{1} }{2.1}=\frac{7-1}{2}=\frac{6}{2}=3$

Pontos das raízes: $(3,0)$ e $(4,0)$