L0rdPh03nix
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if a₀ , a₁ , a₂ , a₃ , are all positive , then 4a₀x³ + 3a₁x² + 2a₂x + a₃ = 0 has at least one root in (-1,0) if
A: a₀ + a₂ = a₁ + a₃ and 4a₀ + 2a₂ > 3a₁ + a₃
B: 4a₀ + 2a₂ < 3a₁ + a₃
C: a₀ + a₂ > a₁ + a₃ and 4a₀ + 2a₂ = 3a₁ + a₃
D: a₀ + a₂ > a₁ + a₃ and 4a₀ + 2a₂ < 3a₁ + a₃

Respuesta :

LammettHash
If [tex]f(x)=4a_0x^3+3a_1x^2+2a_2x+a_3[/tex], then you have

[tex]f(-1)=-4a_0+3a_1-2a_2+a_3[/tex]
[tex]f(0)=a_3[/tex]

By the intermediate value theorem, there will be some number [tex]-1<c<0[/tex] such that [tex]f(c)=0[/tex] (i.e. [tex]f(x)[/tex] will have a root in [tex](-1,0)[/tex]) if you can guarantee that [tex]f(-1)<f(c)<f(0)[/tex] or [tex]f(-1)>f(c)>f(0)[/tex].

Since the coefficients [tex]a_i[/tex] are all positive, then you know right away that [tex]f(0)>0[/tex], so you need to have [tex]f(-1)<0[/tex] in order for there to be such a [tex]c[/tex].

This means you need to have

[tex]-4a_0+3a_1-2a_2+a_3<0\implies 4a_0+2a_2>3a_1+a_3[/tex]

which means (A) must be the answer.

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