Question

2. Indique, entre os valores do conjunto universo, quais são raízes das respectivas equações.
a) x2 - 25 = 0, sendo U = {-5, -2,0, 2,5).
b) x² + 3x = 0, sendo U = {-9, -3,0,3,9}.
c) x2 + 3x - 4 = 0, sendo U = {-4, -1,0, 1,4}.
d) x2 – 2x = 3, sendo U = {-3, -1,0, 1,3}.

Answer 1

Resposta:

a) x² - 25 = 0, sendo U = {-5, -2,0, 2,5).

$(-b±\sqrt{b²-4.a.c})/2.a$

$(0±\sqrt{0²- 4.1.(-25)})/2.1$

$(0±\sqrt{100})/2$

$(0±10)/2$

x1=-10/2 = -5

x2=10/2 = 5

b) x² + 3x = 0, sendo U = {-9, -3,0,3,9}.

$(-b±\sqrt{b²-4.a.c})/2.a$

$(-3±\sqrt{3²-4.1.0})/2.1$

$(-3±\sqrt{9})/2.$

$(-3±3)/2$

x1=(3+3)/2=6/2 = 3

x2=(3-3)/2=0/2 = 0

c) x² + 3x - 4 = 0, sendo U = {-4, -1,0, 1,4}.

$(-b±\sqrt{b²-4.a.c})/2.a$

$(-3±\sqrt{3²-4.1.(-4)})/2.1$

$(-3±\sqrt{9+16})/2$

$(-3±\sqrt{25})/2$

$(-3±5)/2$

x1=(-3+5)/2 = 2/2 = 1

x2=(-3-5)/2 = -8/2 = -4

d) x² – 2x = 3, sendo U = {-3, -1,0, 1,3}.

x² - 2x - 3 = 0

$(-b±\sqrt{b²-4.a.c})/2.a$

$(-(-2)±\sqrt{(-2)²-4.1.(-3)})/2.a$

$(2±\sqrt{(4+12})/2$

$(2±\sqrt{16})/2$

$(2±4)/2$

x1 = (2+4)/2 = 6/2 = 3

x2 = (2-4)/2 = -2/2 = -1