Odpowiedzi

2010-04-15T18:10:05+02:00
X² + (m + 1)x + (- 2m - 6) = 0
Δ = (m + 1)² - 4(- 2m - 6) = m² + 2m + 1 + 8m + 24 = m² + 10m + 25 = (m + 5)²
√Δ = |m + 5|
x₁ = [- m - 1 + |m + 5|]/2
x₂ = [- m - 1 - |m + 5|]/2

a) gdy x₁≠x₂ i x₁,x₂∈<-3;4>

x₁≠x₂ => Δ > 0
(m + 5)² > 0 => m ≠ - 5

- 3 ≤ [- m - 1 + |m + 5|]/2 ≤ 4
- 6 ≤ - m - 1 + |m + 5| ≤ 8
- 5 ≤ - m + |m + 5| ≤ 9
m - 5 ≤ |m + 5| ≤ m + 9

m < - 5
m - 5 ≤ - m - 5 ≤ m + 9
0 ≤ - 2m ≤ 14
0 ≥ m ≥ - 7

m ≥ - 5
m - 5 ≤ m + 5 ≤ m + 9
- 5 ≤ 5 ≤ 9 - zawsze prawda
ostatecznie m ≥ - 7


- 3 ≤ [- m - 1 - |m + 5|]/2 ≤ 4
- 6 ≤ - m - 1 - |m + 5| ≤ 8
- 5 ≤ - m - |m + 5| ≤ 9
m - 5 ≤ - |m + 5| ≤ m + 9

m < - 5
m - 5 ≤ m + 5 ≤ m + 9
- 5 ≤ 5 ≤ 9 - zawsze prawda

m ≥ 5
m - 5 ≤ - m - 5 ≤ m + 9
0 ≤ - 2m ≤ 14
0 ≥ m ≥ - 7
ostatecznie m ≤ 0

czyli:
m ≤ 0 i m ≥ - 7 i m ≠ - 5
m ∈ <- 7; - 5) u (- 5; 0>


b) gdy x₁≠x₂ i 3∈(x₁;x₂)

x₁≠x₂ => Δ > 0
(m + 5)² > 0 => x ≠ - 5

[- m - 1 - |m + 5|]/2 < 3 < [- m - 1 + |m + 5|]/2
- m - 1 - |m + 5| < 6 < - m - 1 + |m + 5|
- |m + 5| < 7 + m < |m + 5|

m < - 5
m + 5 < 7 + m < - m - 5
m + 5 < 7 + m i 7 + m < - m - 5
0 < 2 i m < - 6

m ≥ - 5
- m - 5 < 7 + m < m + 5
- m - 5 < 7 + m i 7 + m < m + 5
- m - 5 < 7 + m i 7 < 5 - sprzeczność

ostatecznie:
m < - 6

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