Question

como resolver a equação de 2°grau
e)6x²-x-2=0
urgenteeeeeeeeee​ ​ ​

Answer 1

✅ Após resolver a equação do segundo grau, concluímos que seu conjunto solução é:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \left\{-\frac{1}{2}, \frac{2}{3}\right\}\:\:\:}}\end{gathered} }$

Seja a equação do segundo grau:

     $\Large\displaystyle\begin{gathered} 6x^{2} - x - 2 = 0\end{gathered}$

Cujos coeficientes são:

             $\Large\begin{cases} a = 6\\b = -1\\c = -2\end{cases}$

Calculando o valor do delta, temos:

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

            $\Large\displaystyle\begin{gathered} = (-1)^{2} - 4\cdot6\cdot(-2)\end{gathered}$

            $\Large\displaystyle\begin{gathered} = 1 + 48\end{gathered}$

            $\Large\displaystyle\begin{gathered} =49\end{gathered}$

       $\Large\displaystyle\begin{gathered} \therefore\:\:\:\Delta = 49\end{gathered}$

Calculando as raízes, temos:

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

            $\Large\displaystyle\text{ \begin{gathered} = \frac{-(-1) \pm\sqrt{49}}{2\cdot6}\end{gathered} }$

            $\Large\displaystyle\begin{gathered} = \frac{1\pm7}{12}\end{gathered}$

Desta forma as raízes são:

    $\LARGE\begin{cases} x' = \frac{1 - 7}{12} = -\frac{6}{12} = -\frac{1}{2}\\x'' = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3}\end{cases}$

✅ Portanto, o conjunto solução é:

          $\Large\displaystyle\text{ \begin{gathered} S = \left\{-\frac{1}{2}, \frac{2}{3}\right\}\end{gathered} }$

 

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Answer 2

$\Large\displaystyle\begin{gathered} \bf{ (I)} \,\blue{\sf Coeficientes\,e\,delta:}\end{gathered}$

$\Large\displaystyle\begin{gathered} \sf{6x^2-x-2=0 } \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \sf{\begin{cases}\sf a=6\\\sf b=-1\\\sf c=-2\end{cases} } \end{gathered} }$

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

$\Large\displaystyle\begin{gathered} \sf{ \Delta=(-1)^2-4\cdot6\cdot(-2)} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf{ \Delta=1+48} \end{gathered}$

$\Large\displaystyle\begin{gathered} \sf{ \Delta=49} \end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \bf{(II) } \,\blue{\sf{Ra\acute{i}zes:}}\end{gathered} }$

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

$\Large\displaystyle\text{ \begin{gathered} \sf{x=\frac{-(-1)\pm\sqrt{49}}{2\cdot6} } \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \sf{ x=\frac{1\pm7}{12}\begin{cases}\sf x_1=\dfrac{1+7}{12}=\dfrac{8}{12}=\dfrac{2}{3}\\\\\sf x_2=\dfrac{1-7}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\end{cases}} \end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \bf{(III) } \,\blue{\sf{Conjunto\, soluc_{\!\!,}\tilde{a}o:}}\end{gathered} }$

$\Large\displaystyle\text{ \begin{gathered} \therefore\green{\underline{\boxed{\boxed{\sf{S=\bigg\{-\frac{1}{2},\frac{2}{3}\bigg\} } }}}}~~(\checkmark)\end{gathered} }$