Question

qual as raizes dessa equação 8x^4 - 20x^2 + 12 = 0

Answer 1

$8 {x}^{4} - 20 {x}^{2} + 12 = 0 \\ 2 {x}^{4} - 5 {x}^{2} + 3 = 0 \\ 2 { ({x}^{2} })^{2} - 5 {x}^{2} + 3 = 0$

Artifício:

${x}^{2} = y$



$2 {y}^{2} - 5y + 3 = 0 \\ \\ delta = 25 - 4 \times 2 \times 3 = 25 - 24 = \\ = 1 \\ \\ y = \frac{5 + 1}{2 \times 2} = \frac{6}{4} = \frac{3}{2} \\ \\ y = \frac{5 - 1}{2 \times 2} = \frac{4}{4} = 1$


Temos que:

${x}^{2} = \frac{3}{2} \\ \sqrt{ {x}^{2} } = \sqrt{ \frac{3}{2} } \\ |x| = \frac{ \sqrt{3} }{ \sqrt{2} } \\ |x| = \frac{ \sqrt{6} }{2} \\ x = \frac{ \sqrt{6} }{2} \: ou \: x = - \frac{ \sqrt{6} }{2} \\ \\ {x}^{2} = 1 \\ \sqrt{ {x}^{2} } = \sqrt{1} \\ |x| = 1 \\ x = 1 \: ou \: x = - 1$


S = {$- \frac{ \sqrt{6} }{2} , -1, 1, \frac{ \sqrt{6} }{2}$}

Answer 2

8x⁴ - 20x² + 12 = 0

8 (x²)² - 20x² + 12 = 0

x² = y

8y² - 20y + 12 = 0

Δ = (-20)² - 4(8)(12)

Δ = 400 - 384

Δ = 16

y = (20 ± 4) ÷ 16

y' = (20 + 4) ÷ 16 = ³/₂

y'' = (20 - 4) ÷ 16 = 1

x² = 1

x = ± √1

x = 1 ou -1

x² = ³/₂

x = ± √³/₂

x = ± √³/√₂

x = ± √⁶/₂

S = {± √⁶/₂; ± 1}