Question

Resolva em R a equação biquadrada:

Answer 1

Resposta:

Solução:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^4 -5x^2 + 4 = 0 \end{array}\right$

Uma equação biquadrada é toda aquela que possui a forma:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf ax^4 + bx^2 + c = 0 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^4 -5x^2 + 4 = 0 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \left ( x^2 \right )^2 -5x^2 + 4 = 0 \end{array}\right$

Fazendo $\sf\textstyle x^2 \: = \: y }$:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf y^2 -5y + 4 = 0 \end{array}\right$

Determinar o Δ:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \sf \displaystyle \Delta = b^2 -\:4ac \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \sf \displaystyle \Delta = (-5)^2 -\:4 \cdot 1 \cdot 4 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \sf \displaystyle \Delta = 25 -\: 16 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \sf \displaystyle \Delta = 9 \end{array}\right$

Determinar as raízes da equação:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf y = \dfrac{-\,b \pm \sqrt{ \Delta } }{2a} = \dfrac{-\,(-5) \pm \sqrt{ 9 } }{2 \cdot 1} = \dfrac{5 \pm 3 }{2} \Rightarrow\begin{cases} \sf y_1 = &\sf \dfrac{5 + 3}{2} = \dfrac{8}{2} = 4 \\\\ \sf y_2 = &\sf \dfrac{5 - 3 }{2} = \dfrac{2 }{2} =1\end{cases}\end{array}\right$

Voltando a condição  $\sf\textstyle x^2 \: = \: y }$:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^2 = y_1 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^2 = 4 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x = \pm \: \sqrt{4} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x = \pm \: 2\end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x_1 = 2\end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x_2 = -\;2\end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^2 = y_2 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x^2 = 1 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x = \pm \: \sqrt{1} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x = \pm \: 1\end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x_3 = 2\end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x_4 = -\:1\end{array}\right$

Portanto, a solução dessa equação biquadrada será {- 2, - 1, 1, 2}.

Explicação passo-a-passo:

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