Dla jakich wartości parametru s równanie x² + 3x + 5s + 2= 0 ma dwa rozwiązania ujemne ? ( założenia)
x² + 3x + 5s + 2 = 0
Musi być:
Δ > 0
x1*x2 = c/a > 0
x1 + x2 = - b/a < 0
x1 + x2 = - 3/1 = - 3 < 0
Mamy więc
Δ = 9 - 4*1*(5s +2) = 9 - 20s - 8 = 1 - 20s
c /a = [5s + 2]/1 = 5s + 2
czyli
1 - 20s > 0
5s + 2 > 0
---------------
-20s > -1
5s > -2
-----------
s < 0,05
s > - 0,4
Odp. s ∈ ( - 0,4 ; 0,05 )
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x² + 3x + 5s + 2 = 0
Musi być:
Δ > 0
x1*x2 = c/a > 0
x1 + x2 = - b/a < 0
x1 + x2 = - 3/1 = - 3 < 0
Mamy więc
Δ = 9 - 4*1*(5s +2) = 9 - 20s - 8 = 1 - 20s
c /a = [5s + 2]/1 = 5s + 2
czyli
1 - 20s > 0
5s + 2 > 0
---------------
-20s > -1
5s > -2
-----------
s < 0,05
s > - 0,4
Odp. s ∈ ( - 0,4 ; 0,05 )
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