Question

06) Sendo S a soma e P o produto das raízes da equação 2x² − 5x − 7 = 0, pode-se afirmar que: A) S − P = 6. B) S + P = 2. C) S ⋅ P = 4. D) S/P= 1 E) S < P

Answer 1

✅ Após as análises concluímos que opção correta quanto às comparações de soma e produto das raízes do segundo grau - equação quadrática - corresponde à:

         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:A\:\:\:}}\end{gathered} }$

Seja a equação do segundo grau:

              $\Large\displaystyle\begin{gathered} 2x^{2} - 5x - 7 = 0\end{gathered}$

Sabemos, pelas relações de Girard, que a soma e o produto das raízes podem ser representados por:

              $\Large\displaystyle\begin{gathered} S = x' + x'' = -\frac{b}{a}\end{gathered}$

                 $\Large\displaystyle\begin{gathered} P = x'\cdot x'' = \frac{c}{a}\end{gathered}$

Então, temos:

• A)

            $\Large\displaystyle\begin{gathered} S - P = -\frac{(-5)}{2} - \frac{(-7)}{2} \end{gathered}$

                           $\Large\displaystyle\begin{gathered} = \frac{5}{2} + \frac{7}{2}\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = \frac{5 + 7}{2}\end{gathered}$

                           $\Large\displaystyle\begin{gathered} = \frac{12}{2}\end{gathered}$

                            $\Large\displaystyle\begin{gathered} = 6\end{gathered}$

               $\Large\displaystyle\begin{gathered} \therefore\:\:\:S - P = 6\end{gathered}$

• B)

              $\Large\displaystyle\begin{gathered} S + P = -\frac{(-5)}{2} + \frac{(-7)}{2} \end{gathered}$

                             $\Large\displaystyle\begin{gathered} = \frac{5}{2} - \frac{7}{2}\end{gathered}$

                             $\Large\displaystyle\begin{gathered} = \frac{5 - 7}{2}\end{gathered}$

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

                              $\Large\displaystyle\begin{gathered} = -1\end{gathered}$

               $\Large\displaystyle\begin{gathered} \therefore\:\:\:S + P \neq 2\end{gathered}$

• C)

                  $\Large\displaystyle\begin{gathered} S\cdot P = -\frac{(-5)}{2}\cdot \frac{-7}{2}\end{gathered}$

                               $\Large\displaystyle\begin{gathered} = -\frac{35}{4}\end{gathered}$

                  $\Large\displaystyle\begin{gathered} \therefore\:\:\:S\cdot P \neq 4\end{gathered}$

• D)

                      $\Large\displaystyle\text{ \begin{gathered} S/P = \frac{-\dfrac{(-5)}{2}}{\dfrac{-7}{2}}\end{gathered} }$

                                  $\Large\displaystyle\begin{gathered} = \frac{5}{\!\diagup\!\!\!\!2}\cdot \frac{\!\diagup\!\!\!\!2}{-7}\end{gathered}$

                                   $\Large\displaystyle\begin{gathered} = -\frac{5}{7}\end{gathered}$

                       $\Large\displaystyle\begin{gathered} \therefore\:\:\:S/P \neq 1\end{gathered}$

• E)

                     $\Large\displaystyle\begin{gathered}S = - \frac{(-5)}{2} = \frac{5}{2} = 2,5\end{gathered}$

                            $\Large\displaystyle\begin{gathered} P = -\frac{7}{2} = -3,5\end{gathered}$

                                        $\Large\displaystyle\begin{gathered} P < S\end{gathered}$

                                   $\Large\displaystyle\begin{gathered} \therefore\:\:\:S \nless P\end{gathered}$

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

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