Question

Achar as raízes das equações: a) x² - x - 20 = 0 b) x² - 3x -4 = 0 c) x² - 8x + 7 = 0. ajuda ​

Answer 1

a) $x^2-x-20=0$

$\triangle=b^2-4.a.c=(-1)^2-4.1.(-20)=1+80=81$

$x_1=\frac{-b+\sqrt{\triangle} }{2a}=\frac{1+\sqrt{81} }{2.1}=\frac{1+9}{2}=\frac{10}{2}=5$

$x_2=\frac{-b-\sqrt{\triangle} }{2a}=\frac{1-\sqrt{81} }{2.1}=\frac{1-9}{2}=\frac{-8}{2}=-4$

$S=\{-4,\ 5\}$

b) $x^2-3x-4=0$

$\triangle=b^2-4.a.c=(-3)^2-4.1.(-4)=9+16=25$

$x_1=\frac{-b+\sqrt{\triangle} }{2a}=\frac{3+\sqrt{25} }{2.1}=\frac{3+5}{2}=\frac{8}{2}=4$

$x_2=\frac{-b-\sqrt{\triangle} }{2a}=\frac{3-\sqrt{25} }{2.1}=\frac{3-5}{2}=\frac{-2}{2}=-1$

$S=\{-1,\ 4\}$

c) $x^2-8x+7=0$

$\triangle=b^2-4.a.c=(-8)^2-4.1.7=64-28=36$

$x_1=\frac{-b+\sqrt{\triangle} }{2a}=\frac{8+\sqrt{36} }{2.1}=\frac{8+6}{2}=\frac{14}{2}=7$

$x_2=\frac{-b-\sqrt{\triangle} }{2a}=\frac{8-\sqrt{36} }{2.1}=\frac{8-6}{2}=\frac{2}{2}=1$

$S=\{1,\ 7\}$