demonstrar que se a equaçao
, 
,
, a e b reais,tiver duas raizes iguais 
sera sempre positivo.
por adauto martins » Qui Jul 29, 2021 17:09
, ,, a e b reais,tiver duas raizes iguais sera sempre positivo.
por adauto martins » Qui Jul 29, 2021 17:38
e 
,onde p'(x),p''(x) sao respectivamente as derivadas primeira e segunda de p(x).
![p'(x)=0\Rightarrow 3x^2+a=0\Rightarrow
x=(+/-)(\sqrt[]{3a}/3)i](/latexrender/pictures/d51913862d80caed67e699abb392d377.png "p'(x)=0\Rightarrow 3x^2+a=0\Rightarrow
x=(+/-)(\sqrt[]{3a}/3)i")
![x=(\sqrt[]{3a}/3)i](/latexrender/pictures/b3b8ab2a9429a156c87c2ce108d07fde.png "x=(\sqrt[]{3a}/3)i")
![x=-(\sqrt[]{3a}/3)i](/latexrender/pictures/a02c91f52ccf75bf9c41371c415ec504.png "x=-(\sqrt[]{3a}/3)i")
i)=({(\sqrt[]{3}a/3)i})^{3}+a(\sqrt[]{3}a/3)i+b=0](/latexrender/pictures/1a0dfe20eed241d1f1af4fc130abf50d.png "p(\sqrt[]({3}a/3)i)=({(\sqrt[]{3}a/3)i})^{3}+a(\sqrt[]{3}a/3)i+b=0")
{i})^{2}.(\sqrt[]{3a}/3)i+a(\sqrt[]{3a}/3)i+b=0...
...=4a^3+27b^2\Rightarrow 4a^3=-27b^2\prec 0\Rightarrow a^3=a^2.a\prec 0
\Rightarrow a\prec 0](/latexrender/pictures/0392bf6c134ee57e17697c566af2f6a6.png "\sqrt[](({3a}/3){i})^{2}.(\sqrt[]{3a}/3)i+a(\sqrt[]{3a}/3)i+b=0...
...=4a^3+27b^2\Rightarrow 4a^3=-27b^2\prec 0\Rightarrow a^3=a^2.a\prec 0
\Rightarrow a\prec 0")
![p(-\sqrt[]{3a}/3)=0\Rightarrow 4a^3+27p^2=0](/latexrender/pictures/b37b9596ec387081db874b9523874f9b.png "p(-\sqrt[]{3a}/3)=0\Rightarrow 4a^3+27p^2=0")
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![\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}](/latexrender/pictures/981987c7bcdf9f8f498ca4605785636a.png "\frac{\sqrt[]{\sqrt[4]{8}+\sqrt[]{\sqrt[]{2}-1}}-\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}-1}}}{\sqrt[]{\sqrt[4]{8}-\sqrt[]{\sqrt[]{2}+1}}}")
![\sqrt[]{2}](/latexrender/pictures/f21662d1cabab6e8b273a4b6f1cd663a.png "\sqrt[]{2}")
(dica : igualar a expressão a 
e elevar ao quadrado os dois lados)