При каких значениях параметра p отношение корней уравнения x^2 + 2px+1=0 равно 9?

Решение

x² + 2px + 1 = 0

D = 4p² - 4*1*1 = 4 (p² - 1)

x₁ = (- 2p - 2√ (p² - 1)) / 2

x₁ = - p - √ (p² - 1)

x₂ = (- 2p + 2√ (p² - 1)) / 2

x₂ = - p + √ (p² - 1)

x₁ : x₂ = 9

[ - p - √ (p² - 1) ] : [ - p + √ (p² - 1) ] = 9

- p - √ (p² - 1) = 9 * [ - p + √ (p² - 1) ]

- p - √ (p² - 1) = - 9p + 9√ (p² - 1)

- 9p + 9√ (p² - 1) + p + √ (p² - 1) = 0

10√ (p² - 1) = 8p

[10√ (p² - 1) ]² = (8p) ²

100 (p² - 1) = 64p²

100p² - 100 - 64p² = 0

36p² = 100

p² = 100/36

p₁ = - 10/6 = - 5/3 = - 1 (2/3)

p₂ = 10/6 = 1 (2/3)