Question

Помогите решить пж:
x^4+10x^3-125x-54=0

Answer 1

x⁴ + 10x³ - 125x - 54 = o

x⁴ + 10x³ + 25x² - 25x² - 125x - 54 = 0

(x⁴ + 10x³ + 25x²) - (25x² + 125x) - 54 = 0

(x² + 5x)² - 25(x² + 5x) - 54 = 0

Сделаем замену : x² + 5x = m

m² - 25m - 54 = 0

m₁ = 27        m₂ = - 2

1) x² + 5x = 27

x² + 5x - 27 = 0

D = 5² - 4 * (- 27) = 25 + 108 = 133

$x_{1}=\frac{-5+\sqrt{133} }{2}\\\\x_{2}=\frac{-5-\sqrt{133} }{2}$

2)x² + 5x = - 2

x² + 5x + 2 = 0

D = 5² - 4 * 2 = 25 - 8 = 17

$x_{3}=\frac{-5+\sqrt{17} }{2}\\\\x_{4}=\frac{-5-\sqrt{17} }{2}$