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m x² - 3(m +1 ) x + m = 0
Δ = [ -3(m +1)]² - 4*m*m = 9*(m²+2m +1) - 4 m² =
= 9 m² + 18 m + 9 - 4 m² = 5 m² + 18 m + 9
Aby równanie dane nie miało rozwiązań rzeczywistych
Δ musi być ujemna, zatem
5 m² + 18 m + 9 < 0
Δ1 = 18² -4*5*9 = 324 - 180 = 144
√Δ1 = 12
m1 = [-18 - 12]/10 = -30/10 = -3
m2 = [-18 + 12]/10 = -6/10 = - 0,6
a = 5 > 0 zatem 5 m² + 18 m + 9 < 0 <=> m ∈ ( -3 ; -0,6 )
Odp. m ∈ ( -3 ; -0,6 )