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Równanie ma 2 rózne pierwiastki dodatnie gdy:
1) Δ>0
2) x1+x2>0
3)x1*x2>0
Ad1
Δ = (-2(m-2))^2 - 4*1*(m^2-2m-3) = 4(m^2 - 4m + 4) - 4m^2 + 8m + 12 =
4m^2-16m+16 - 4m^2 + 8m + 12 = -8m + 28
-8m+28 >0
-8m > -28 |:(-8)
m < 3,5
m ∈ (-∞, 3,5)
Ad 2
x1 + x2 = -b/a (ze wzorow Viete'a)
x1 + x2 = 2(m-2)/1 = 2m - 4
2m - 4 >0
2m > 4 |:2
m>2
m∈ (2 , +∞)
Ad 3
x1 *x2 = c/a
x1*x2 = (m^2-2m-3)/1 = m^2-2m-3
m^2 - 2m - 3 >0
Δ = (-2)^2 - 4*1*(-3) = 4 + 12 = 16
√Δ = 4
m1 = (2-4)/2 = -1 m2 = (2+4)/2 = 3
m∈(-∞ , -1 ) ∨ (3 , +∞)
Część wspólna rozwiązań z wszystkich trzech warunków:
m ∈ (3 , 3,5)