Question

Resolva a equação: x³ - x = 2 , utilizando a fórmula de Cardano

Answer 1

Resposta:

$\boxed{\bold{x_1=\sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \sqrt[3]{\dfrac{9-\sqrt{78}}{9}}}}$

$\boxed{\bold{x_2=-\dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9-\sqrt{78}}{9}}}}\\\\\\ \boxed{\bold{x_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9-\sqrt{78}}{9}}}}}$

Explicação passo-a-passo:

Olá, boa noite.

Devemos resolver a equação algébrica de grau 3 $x^3-x=2$ utilizando a fórmula de Cardano.

Primeiro, subtraia 2 em ambos os lados da equação

$x^3-x-2=0$

Seja uma equação de grau 3 com coeficientes reais $ax^3+bx^2+cx+d=0$, com $a\neq0$. A fórmula de Cardano prevê a existência de três raízes, dadas por:

$x_1=-\dfrac{b}{3a}+\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+ \sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\\\\\\ x_2=-\dfrac{b}{3a}+\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}$

$x_3=-\dfrac{b}{3a}+\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}$

Tal que $q=\dfrac{d}{a}-\dfrac{b\cdot c}{3\cdot a^2}+\dfrac{2b^3}{27a^3}$ e $p=\dfrac{c}{a}-\dfrac{b^2}{3a^2}$.

Observe que nossa equação está incompleta, apresentando somente os coeficientes $a=1,~c=-1$ e $d=-2$. Essencialmente, consideramos $b=0$.

Substituindo os valores dos coeficientes para calcular $q$ e $p$, temos:

$q=\dfrac{-2}{1}-\dfrac{0\cdot (-1)}{2}+\dfrac{2\cdot 0^3}{27\cdot 1^3}$ e $p=\dfrac{-1}{1}-\dfrac{0^2}{3\cdot 1^2}$

Multiplique e some os valores

$q=-2$ e $p=-1$

Substituindo os valores na fórmula de Cardano, temos as soluções

$x_1=\sqrt[3]{-\dfrac{-2}{2}+\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}+ \sqrt[3]{-\dfrac{(-2)}{2}-\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}\\\\\\ x_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{(-2)}{2}+\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot\sqrt[3]{-\dfrac{(-2)}{2}-\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}$$x_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-\dfrac{(-2)}{2}+\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{-\dfrac{(-2)}{2}-\sqrt{\dfrac{(-2)^2}{4}+\dfrac{(-1)^3}{27}}}$

Calculando as potências e multiplicando os valores, temos

$x_1=\sqrt[3]{1+\sqrt{1+\dfrac{-1}{27}}}+ \sqrt[3]{-1-\sqrt{1+\dfrac{-1}{27}}}\\\\\\ x_2=\dfrac{-1+i\sqrt{3}}{2}\cdot \sqrt[3]{1+\sqrt{1+\dfrac{-1}{27}}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{-1-\sqrt{1+\dfrac{-1}{27}}}$

$x_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{1+\sqrt{1+\dfrac{-1}{27}}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{1-\sqrt{1+\dfrac{-1}{27}}}$

Somando os valores, temos

$x_1=\sqrt[3]{1+\sqrt{\dfrac{26}{27}}}+ \sqrt[3]{-1-\sqrt{\dfrac{26}{27}}}\\\\\\ x_2=-\dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{1+\sqrt{\dfrac{26}{27}}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{1-\sqrt{\dfrac{26}{27}}}\\\\\\ x_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{1+\sqrt{\dfrac{26}{27}}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{1-\sqrt{\dfrac{26}{27}}}$

Podemos simplificar as raízes quadradas, sabendo que $\sqrt{\dfrac{m}{n}}=\dfrac{\sqrt{m}}{\sqrt{n}}$ e $27=3^3$

$x_1=\sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \sqrt[3]{\dfrac{9-\sqrt{78}}{9}}\\\\\\ x_2=-\dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \dfrac{-1-i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9-\sqrt{78}}{9}}\\\\\\ x_3=\dfrac{-1-i\sqrt{3}}{2}\cdot \sqrt[3]{\dfrac{9+\sqrt{78}}{9}}+ \dfrac{-1+i\sqrt{3}}{2}\cdot\sqrt[3]{\dfrac{9-\sqrt{78}}{9}}$

Utilizando o auxílio de uma calculadora científica, observamos que

$x_1\approx 1.52137971\\\\\\ x_2\approx-0.76069 - 0.85787 i\\\\\\ x_3\approx-0.76069 + 0.85787 i$

Estas são as soluções para esta equação.