Question

Resolver, em R, a equação 2x² – 4x – 16 = 0

Answer 1

Resposta:

$\Large\checkmark~\Large\text{\bf{x_{1} ~=~4}}\\\\\checkmark~\Large\text{\bf{x_{2} ~=~ - ~2}}$

Cálculo:

$\boxed{\Large\text{\bf{2x^2 ~-~4x~-~16~=~0}}}$

$\large\text{}\begin{cases}a~=~ \text{2} \\b~=~\text{ - ~4}\\c~=~\text{ - ~16}\end{cases}$

$\Large\text{\bf{\Delta ~=~b^2 ~-~4~.~a~.~c}}$

$\large\text{\Delta ~=~(-4)^2 ~-~4~.~2~.~(-16)}\\\\\large\text{\Delta ~=~16~-~4~.~2~.~(-16)}\\\\\large\text{\Delta ~=~16~-~8~.~(-16)}\\\\\large\text{\Delta ~=~16~+~128}\\\\\large\text{\Delta ~=~144}$

$\Large\text{x~=~ \bf{\dfrac{-b~\pm~\sqrt{\Delta}}{2~.~a}} }$

$\large\text{x~=~ \dfrac{-(-4)~\pm~\sqrt{144}}{2~.~2} }\\\\\\\large\text{x~=~ \dfrac{+4~\pm~12}{4} }\\\\\\\large\text{x_{1} ~=~ \dfrac{+4~+~12}{4} }\\\\\\\large\text{x_{1} ~=~ \dfrac{16}{4} }\\\\\\\boxed{\boxed{\large\text{x_{1} ~=~4}}}\\\\\\\large\text{x_{2} ~=~ \dfrac{+4~-~12}{4} }\\\\\\\large\text{x_{2} ~=~ \dfrac{-~8}{4} }\\\\\\\boxed{\boxed{\large\text{x_{2} ~=~ - ~2}}}$

Answer 2

$\LARGE\text{ \underline {\sf Ol\acute{a}{,}\ boa\ tarde!} }$

              $\searrow$

☛  $\large\text{ \underline{\sf Equac_{\!\!\!,}\tilde{a}o\ do\ 2^o\ Grau.} }$ ☃️

➡  $\Large \sf Como\ calcular\ uma\ equac_{\!\!\!,}\tilde{a}o\ do\ 2^o\ Grau\ ?$

$\large\sf Para\ calcular\ uma\ equac_{\!\!\!,}\tilde{a}o\ do\ 2^o\ grau,temos\ 2\ f\acute{o}rmulas.$  

             $\swarrow$

➡️  $\LARGE\text{ \underline{\sf S\tilde{a}o\ elas}: }$

✏️ $\large\text{ \underline{\sf F\acute{o}rmula\ do\ discriminante}: }$

  $\LARGE\boxed{\sf \Delta= b^2 -4\cdot a\cdot b }$

✏️ $\large\text{ \underline{\sf F\acute{o}rmula\ de\ bhaskara}: }$

   $\LARGE\boxed{\sf \dfrac{-b\ \pm\ \sqrt{\Delta} }{2\cdot a} }$

➡️  $\large\sf Mas\ antes\ de\ aplicar\ as\ f\acute{o}rmulas{,}\ devemos\ achar\ os\ valores$

$\large\sf de\ (a{,}\ b{,}\ c) .$

                      $\searrow$

✏️ $\Large\text{ \underline{\sf Descobrindo\ os\ valores\ de\ (a{,}\ b{,}\ c)\ da\ sua\ tarefa}: }$

           

  $\Longrightarrow\ \large\sf 2x^2 - 4x - 16 = 0 \large\begin{cases}\sf a= 2\\ \sf b= -4\\ \sf c=-16\end{cases}$

$\Large\text{ \underline{\sf Pronto{,}\ encontramos\ os\ termos\ (a{,}\ b{,}\ c).} }$

           $\Downarrow$

$\LARGE\boxed{\boxed{\begin{array}{l}\sf a= 2\\ \sf b= -4\\ \sf c= -16 \end{array}}}$

✏️ $\Large\text{ \underline{\sf Agora{,}\ podemos\ aplicar\ a\ f\acute{o}rmula\ do\ discriminante.} }$

$\large \sf \Delta= (-4)^2 -4\cdot 2\cdot (-16) \\\\ \large \sf \Delta= 16 -8\cdot (-16) \\\\ \large \sf \Delta= 16 + 128\\\\ \large \sf \Delta= 144$

$\Large\text{ \underline{\sf Pronto{,}\ encontramos\ o\ discriminante.} }$

             $\Downarrow$

 $\LARGE\boxed{\boxed{\begin{array}{l}\sf \Delta= 144\end{array}}}$

✏️ $\Large\text{ \underline{\sf Agora{,}\ podemos\ aplicar\ a\ f\acute{o}rmula\ de\ bhaskara.} }$

  $\large\text{ \sf \dfrac{-(-4)\ \pm\ \sqrt{144} }{4} } \\\\\\\large \sf \dfrac {4\ \pm\ 12 }{4} \\\\\\\large \sf \dfrac {16}{4} = 4$

$\Large\text{ \underline{\sf Pronto{,}\ encontramos\ o\ primeiro\ valor\ de\ x.} }$

           $\Downarrow$

 $\LARGE\boxed{\boxed{\begin{array}{l}\sf x^1 = 4\end{array}}}$

✏️ $\Large \underline{\sf Encontrando\ o\ segundo\ valor\ de\ x.}$

 $\large\text{ \sf \dfrac{-(-4)\ \pm\ \sqrt{144} }{4} } \\\\\\\large \sf \dfrac {4\ \pm\ 12 }{4} \\\\\\\large \sf -\dfrac {8}{4} = -2$

$\Large\text{ \underline{\sf Pronto{,}\ encontramos\ o\ segundo\ valor\ de\ x.} }$

             $\Downarrow$

 $\LARGE\boxed{\boxed{\begin{array}{l}\sf x^2 = -2\end{array}}}$

⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶⎶

☃️ $\Large\text{ \underline{\sf Portanto\ a\ soluc_{\!\!\!,}\tilde{a}o\ dessa\ equac_{\!\!\!,}\tilde{a}o\ do\ 2^o\ Grau\ \acute{e}}: }$

                   $\Downarrow$

   $\LARGE\boxed{\boxed{\begin{array}{l}\sf S = \left\{-2{,}\ 4\right\}\end{array}}}$

$\LARGE\begin{matrix} \underbrace{ \sf By: Pedro } \end{matrix}$