Question

(x²-2x-3)²-2(x²-2x-3)-4=0, решите с полным объяснением.

Answer 1

ax² + bx + c  = 0

D = b² - 4ac

x₁₂ = (-b ± √D) / 2a

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(x²-2x-3)²-2(x²-2x-3)-4=0

замена (x²-2x-3) = y

y² - 2y - 4 = 0

D = 4 + 16 = 20

√20 = 2√5

y₁₂ = (2 +- 2√5)/2 = 1 ± √5

1. y = 1 + √5

x²-2x-3 = 1 + √5

x²-2x-4 - √5 = 0

D = 4 + 4*(4 + √5) = 20 + 4√5

x₁₂ = (2 +- 2√(5 + √5))/2 = 1 ± √(5 + √5)

2.  y = 1 - √5

x²-2x-3 = 1 - √5

x²-2x-4 + √5 = 0

D = 4 + 4*(4 - √5) = 20 - 4√5

x₃₄ = (2 +- 2√(5 -√5))/2 = 1 ± √(5 - √5)

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x = $1 +- \sqrt{5 + \sqrt{5} }\\ 1 +- \sqrt{5 - \sqrt{5} }$