Odpowiedź:
Szczegółowe wyjaśnienie:
x²+2x+m-1=0
Δ=2²-4(m-1) = 4 -4m +4 = -4(m-2)
1) warunek 1 -> Δ>0
-4(m-2)>0
m-2<0
m<2
m∈(-∞,2)
2) warunek 2 -> |x₁|+|x₂|≤3 |²
(|x₁|+|x₂|)²≤3²
|x₁|² +2|x₁x₂|+|x₂|² ≤ 9
x₁²+x₂²+2|x₁x₂| ≤ 9
(x₁+x₂)²-2x₁x₂+2|x₁x₂|≤9
|x₁x₂| = |c/a| = |m-1|
-> |x₁x₂| dla m ≥ 1 -> m-1
b²/a² -2c/a +2(m-1) ≤9
4-2(m-1)+2(m-1)≤9
4≤9, m∈[1, ∞]
-> |x₁x₂| dla m<1 -> 1-m
4-2(m-1) +2(1-m)≤9
4-2m+2+2-2m≤9
8-4m≤9
-4m≤1
m≥ -1/4 -> m∈[-1/4, ∞)
m∈(-∞,2) ∩ m∈[1, ∞] ∩ m∈[-1/4, ∞)) -> m∈[-1/4,2)
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Odpowiedź:
Szczegółowe wyjaśnienie:
x²+2x+m-1=0
Δ=2²-4(m-1) = 4 -4m +4 = -4(m-2)
1) warunek 1 -> Δ>0
-4(m-2)>0
m-2<0
m<2
m∈(-∞,2)
2) warunek 2 -> |x₁|+|x₂|≤3 |²
(|x₁|+|x₂|)²≤3²
|x₁|² +2|x₁x₂|+|x₂|² ≤ 9
x₁²+x₂²+2|x₁x₂| ≤ 9
(x₁+x₂)²-2x₁x₂+2|x₁x₂|≤9
|x₁x₂| = |c/a| = |m-1|
-> |x₁x₂| dla m ≥ 1 -> m-1
(x₁+x₂)²-2x₁x₂+2|x₁x₂|≤9
b²/a² -2c/a +2(m-1) ≤9
4-2(m-1)+2(m-1)≤9
4≤9, m∈[1, ∞]
-> |x₁x₂| dla m<1 -> 1-m
(x₁+x₂)²-2x₁x₂+2|x₁x₂|≤9
4-2(m-1) +2(1-m)≤9
4-2m+2+2-2m≤9
8-4m≤9
-4m≤1
m≥ -1/4 -> m∈[-1/4, ∞)
m∈(-∞,2) ∩ m∈[1, ∞] ∩ m∈[-1/4, ∞)) -> m∈[-1/4,2)