Question

(x-3)^4-5(x-3)^2-14=0

Answer 1

Сделаем замену: пусть (x-3)² = t, тогда:
t² -5t - 14 = 0
D = 25 + 56 = 81 = 9²
t₁ = $frac{5 + 9}{2} = 7$
t₂ = $frac{5 - 9}{2} = -2$

1) (x - 3)² = 7
x² - 6x + 9 - 7 = 0
x² - 6x + 2 = 0
D = 36 - 8 = 28 = 2√7
x₁ = 3 + √7
x₂ = 3 - √7

2) (x - 3)² = -2
x² - 6x + 9 + 2 = 0
x² - 6x + 11 = 0
D = 36 - 44 = ∅