Question

решение пожалуйста x^4-20x^2+64=0

Answer 1

${x}^{4} - 20x {}^{2} + 64 = 0 \\ (y = {x}^{2} ) \\ {y}^{2} - 20y + 64 = 0 \\ d = b {}^{2} - 4ac = 400 - 256 = 144 \\ y_{1.2} = \frac{ - b + - \sqrt{d} }{2a} = \frac{20 + - \sqrt{144} }{ 2 } = 16 \: \: \: \: \: or \: \: \: \: \: 4 \\ \\ (y = {x}^{2} ) \\ {x}^{2} = 16 \\ x {}^{2} { = 4} \\ x = + - 2 \: \: \: \: or \: \: \: \: + - 4$
Ответ : -4 ; -2 ; 2 ; 4

Answer 2

х⁴ - 20х² + 64 = 0

х² = t.

t² - 20t + 64 = 0.
D = 400 - 4 × 1 × 64 = 400 - 256 = 144 > 0. 12.
х1 = 20 + 12 / 2 = 32/2 = 16.
х2 = 20 - 12 / 2 = 8/2 = 4.

х² = 16
х = ± 4.

х² = 4
х = ± 2.

Удачи)))))