Question

x4-12x2+32=0
preciso da conta inteira

Answer 1

$x^4-12x^2+32=0$

$a=1\quad b=-12\quad c=32$

$\Delta=b^2-4\cdot a\cdot c$

$\Delta=(-12)^2-4\cdot 1\cdot32$

$\Delta=144-128$

$\Delta=16$

$x=\pm\sqrt{\dfrac{-b\pm\sqrt\Delta}{2\cdot a}}$

$x=\pm\sqrt{\dfrac{-(-12)\pm\sqrt{16}}{2\cdot1}}$

$x=\pm\sqrt{\dfrac{12\pm4}{2}}\begin{cases}x' =+\sqrt{\dfrac{12+4}{2}}=+\sqrt{8}=+2\sqrt2\\\\x''=-\sqrt{\dfrac{12+4}{2}}=-\sqrt8=-2\sqrt2\\\\x'''=+\sqrt{\dfrac{12-4}{2}}=+\sqrt4=+2\\\\x''''=-\sqrt{\dfrac{12-4}{2}}=-\sqrt4=-2\end{cases}$

$\red{\boldsymbol{ S=\big\{ -2\sqrt2;\;-2;\;+2;\;+2\sqrt2\big\}}}$

Answer 2

✅ Após resolver os cálculos, concluímos que o conjunto solução da referida equação do biquadrada é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \{-2\sqrt{2},\,-2,\,2,\,2\sqrt{2}\}\:\:\:}}\end{gathered} }$

   

Seja a equação biquadrada:

       $\Large\displaystyle\begin{gathered} x^{4} - 12x^{2} + 32 = 0\end{gathered}$

Cujos coeficientes são:

               $\Large\begin{cases} a = 1\\b = -12\\c = 32\end{cases}$    

Resolvendo a equação biquadrada, devemos utilizar a seguinte estratégia:

    $\Large\displaystyle\text{ \begin{gathered} x = \pm\sqrt{\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}\end{gathered} }$

         $\Large\displaystyle\text{ \begin{gathered} = \pm\sqrt{\frac{-(-12) \pm\sqrt{(-12)^{2} - 4\cdot1\cdot32}}{2\cdot1}}\end{gathered} }$

          $\Large\displaystyle\text{ \begin{gathered} = \pm\sqrt{\frac{12\pm\sqrt{144 - 128}}{2}}\end{gathered} }$

          $\Large\displaystyle\text{ \begin{gathered} = \pm\sqrt{\frac{12 \pm \sqrt{16}}{2}}\end{gathered} }$

           $\Large\displaystyle\text{ \begin{gathered} = \pm\sqrt{\frac{12 \pm 4}{2}}\end{gathered} }$

Desta forma, obtemos as seguintes raízes:

 $\Large\begin{cases} x' = -\sqrt{\frac{12 + 4}{2}} = -\sqrt{\frac{16}{2}} = -\sqrt{8} = -2\sqrt{2}\\x'' = -\sqrt{\frac{12 - 4}{2}} = -\sqrt{\frac{8}{2}} = -\sqrt{4} = -2\\x''' = \sqrt{\frac{12 - 4}{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\\x'''' = \sqrt{\frac{12 + 4}{2}} = \sqrt{\frac{16}{2}} = \sqrt{8} = 2\sqrt{2}\end{cases}$

✅ Portanto, o conjunto solução da referida equação é:

           $\Large\displaystyle\text{ \begin{gathered} S = \{-2\sqrt{2},\,-2, \,2,\,2\sqrt{2}\}\end{gathered} }$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$