Question

 determine o zeros das funções

A)x^4-4x^2+4

Answer 1

Explicação passo-a-passo:

$condicao \: x {}^{2} = t \\ x {}^{4} - 4x {}^{2} + 4 = 0 \\ t {}^{2} - 4t + 4 = 0 \\ t = \frac{ -( - 4 ) + \sqrt{( - 4) {}^{2} - 4 \times 1 \times 4 } }{2 \times 1} \\ t = \frac{ -( - 4 )+ \sqrt{16 - 16} }{2} \\ t = \frac{ 4 + 0}{2} \\ t 1= 2 \\ t2 = 2 \\ x {}^{2} = 2 \\ x 1= \sqrt{2} \\ x 2= - \sqrt{2} \\ x 3= \sqrt{2} \\ x 4= - \sqrt{2}$

Answer 2

✅ Após resolver a função biquadrada, concluímos que suas raízes são:

     $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \Large\begin{cases}\tt x' =\sqrt{2}\\\tt x'' = \sqrt{2}\\x''' = -\sqrt{2}\\x'''' = \sqrt{2} \end{cases}\:\:\:}}\end{gathered} }$

Seja a função biquadrada:

        $\Large\displaystyle\begin{gathered} f(x) = x^{4} - 4x^{2} + 4\end{gathered}$

Esta função é a mesma:

      $\Large\displaystyle\begin{gathered} f(x) = (x^{2})^{2} - 4x^{2} + 4\end{gathered}$

Fazendo:

                     $\Large\displaystyle\begin{gathered} x^{2} = \lambda\end{gathered}$

Temos:

        $\Large\displaystyle\begin{gathered} f(\lambda) = \lambda^{2} - 4\lambda + 4\end{gathered}$    

Calculando o valor do delta, temos:

        $\Large\displaystyle\begin{gathered} \Delta = b^{2} - 4ac\end{gathered}$

             $\Large\displaystyle\begin{gathered} = (-4)^{2} - 4\cdot1\cdot4\end{gathered}$

             $\Large\displaystyle\begin{gathered} = 16 - 16\end{gathered}$

             $\Large\displaystyle\begin{gathered} = 0\end{gathered}$

Obtendo o valor de "λ":

             $\Large\displaystyle\text{ \begin{gathered} \lambda = \frac{-b\pm\sqrt{\Delta}}{2a}\end{gathered} }$

                 $\Large\displaystyle\text{ \begin{gathered} = -\frac{(-4) \pm\sqrt{0}}{2\cdot 1}\end{gathered} }$

                 $\Large\displaystyle\begin{gathered} = \frac{4}{2}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} = 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} \therefore\:\:\lambda = 2\end{gathered}$

Agora devemos calcular os valores das raízes. Desta forma fazemos:

                        $\Large\displaystyle\begin{gathered} x^{2} = \lambda\end{gathered}$

                        $\Large\displaystyle\begin{gathered} x^{2} = 2\end{gathered}$

                        $\Large\displaystyle\begin{gathered} x = \pm\sqrt{2}\end{gathered}$

✅ Portanto, as raízes da função biquadrada são:

    $\Large\displaystyle\begin{gathered} x' = -\sqrt{2},\:\:x'' = \sqrt{2},\:\:x''' = -\sqrt{2}\:\:e\:\:x'''' = \sqrt{2}\end{gathered}$

             

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Veja a solução gráfica representada na figura: