Question

A)x⁴ -8x² +12=0
B) 4x⁴ -28x² +48=0

Answer 1

Olá!!

Resolução!!

a)

x⁴ - 8x² + 12 = 0
( x² )² - 8x² + 12 = 0

x² = k

k² - 8k + 12 = 0

a = 1, b = - 8, c = 12

∆ = b² - 4ac
∆ = ( - 8 )² - 4 • 1 • 12
∆ = 64 - 48
∆ = 16

k = - b ± √∆ / 2a
k = - ( - 8 ) ± √16 / 2 • 1
k = 8 ± 4/2
k' = 8 + 4/2 = 12/2 = 6
k" = 8 - 4/2 = 4/2 = 2

k = 6 e k = 2

x² = k

x² = 6 → x = ± √6
x² = 2 → x = ± √2

S = { - √2, - √6, √2, √6 }

b)

4x² - 28x² + 48 = 0
4 ( x² )² - 28x² + 48 = 0

x² = k

4k² - 28k + 48 = 0 ÷ 4
k² - 7k + 12 = 0

a = 1. b = - 7, c = 12

∆ = b² - 4ac
∆ = ( - 7 )² - 4 • 1 • 12
∆ = 49 - 48
∆ = 1

k = - b ± √∆ / 2a
k = - ( - 7 ) ± √1 / 2 • 1
k = 7 ± 1/2
k' = 7 + 1/2 = 8/2 = 4
k" = 7 - 1/2 = 6/2 = 3

k = 4 e k = 3

x² = k

x² = 4 → x = ± √4 → x ± 2
x² = 3 → x = ± √3

S = { - √3, - 2, √3, 2 }

Espero ter ajudado!!