ma poate ajuta careva ?
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Răspuns:
(2m+1)x^2+3x+1=0; a = 2m+1 ; b = 3 ; c = 1
a) si b) Sa aiba radacini reala, adica delta > 0
Delta = b^2- 4ac = 9-4(2m+1) = 9-8-8m > 0 => -8m + 1 > 0
-8m > -1 => 8m < 1 => m > 1/8
c) Delta = 0, adica x1 = x2
b^2-4ac = 0 => -8m+1 = 0 => 8m = 1 => m = 1/8
d) Delta < 0 => m < 1/8
Explicație pas cu pas: