x^4 + 4x^3 + ax^2 + 4x + 1 =0
carui interval apartine a a.i. toate radacinile sunt reale


Răspuns :

x⁴+4x³+ax²+4x+1=0  l: x²
x²+4x+a+4/x+1/x²=0
(x²+1/x²) + 4(x+1/x) + a = 0
Notam (x+1/x)=t    ⇒    (x²+1/x²) = t²-2
t²-2+4t+a=0
t²+4t+a-2=0
Δ=b² - 4ac = 16 - 4a +8 > 0    ⇒    4a > 24   ⇒   a > 6   ⇒   a ∈ (6, +∞)

x⁴+4x³+6x²+4x+1=0