Question

Resolva as equações modulares:
a) (x - 2) = 4
b) (x - 6) = 10
c) (x + 8) = 12
d) (x² - 5 . x + 6) = 2
e) (x² - 8 . x + 13) = 1

Answer 1

Resposta:

Explicação passo-a-passo:

a)

|x - 2| = 4

Conjunto A

x - 2 = 4

x = 6

Conjunto B

-x + 2 = 4

x = -2

A∪B = {-2  6}

d)

|x² - 5x + 6| = 2

Conjunto A

x² - 5x + 6 = 2

x² - 5x + 4= 0

(x -4)(x -1) = 0

x - 4 = 0 ⇒ x' = 4

x - 1 = 0 ⇒ x'' = 1

A = { 1  4}

Conjunto B

-x² + 5x - 6 = 2

-x² + 5x - 8 = 0

x² - 5x + 8 = 0

Δ < 0

B = ∅

A∪B = {1  4}

as equações "b" "c" e "e" ficam para seu treinamento... basta usar o mesmo procedimento da "a" para resolver as "b'' e "c" e usar o procedimento da "d" para resolver a "e"

Answer 2

Resposta:

a) (x - 2) = 4

x - 2 = 4

x = 4 + 2

x = 6

b) (x - 6) = 10

x - 6  = 10

x = 10 + 6

x = 16

c) (x + 8) = 12

x + 8 = 12

x = 12 - 8

x = 4

d) (x² - 5 . x + 6) = 2

x² - 5x + 6 = 2

x² - 5x + 6 - 2 = 0

x² - 5x + 4 = 0

a = 1

b = -5

c = 4

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-5) +- \sqrt{(-5)^{2} - 4 * 1 * 4} }{2*1}\\\\x = \frac{5+-\sqrt{25-16} }{2}\\\\x = \frac{5+-\sqrt{9} }{2}\\\\x = \frac{5+-3}{2}\\\\x' = \frac{5+3}{2}=\frac{8}{2} =4\\\\x'' = \frac{5-3}{2}=\frac{2}{3} =1$

e) (x² - 8 . x + 13) = 1

x² - 8x + 13 = 1

x² - 8x + 13 - 1 = 0

x² - 8x + 12 = 0

a = 1

b = -8

c = 12

$x = \frac{-(b) +- \sqrt{(b)^{2} - 4 * a * c} }{2*a}\\\\x = \frac{-(-8) +- \sqrt{(-8)^{2} - 4 * 1 * 12} }{2*1}\\\\x = \frac{8+-\sqrt{64-48} }{2}\\\\x = \frac{8+-\sqrt{16} }{2}\\\\x = \frac{8+-4}{2}\\\\x' = \frac{8+4}{2}=\frac{12}{2}=6 \\\\x'' = \frac{8-4}{2}=\frac{4}{2} =2$