Question

f(x)=2x-9 e g(x)=x2+5x+3. A soma dos valores absolutos das raizes da equacao f(g(x))=g(x) e igual a

Answer 1

✅ Após resolver os cálculos, concluímos que a soma dos valores absolutos das raízes da referida equação é:

                         $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S_{a} = 7\:\:\:}}\end{gathered} }$

Sejam as funções polinomiais:

                   $\Large\begin{cases} f(x) = 2x - 9\\g(x) = x^{2} + 5x + 3\end{cases}$

Sabemos que para calcular a composição f(g(x)) devemos substituir a variável "x" da função "f(x)" por "g(x)". então temos:

                            $\Large\displaystyle\begin{gathered} f(g(x)) = g(x)\end{gathered}$

                    $\Large\displaystyle\begin{gathered} 2\cdot g(x) - 9 = g(x)\end{gathered}$

 $\Large\displaystyle\begin{gathered} 2\cdot(x^{2} + 5x + 3) - 9 = x^{2} + 5x + 3\end{gathered}$

      $\Large\displaystyle\begin{gathered} 2x^{2} + 10x + 6 - 9 = x^{2} + 5x + 3\end{gathered}$

               $\Large\displaystyle\begin{gathered} 2x^{2} + 10x - 3 = x^{2} + 5x + 3\end{gathered}$

           $\Large\displaystyle\begin{gathered} 2x^{2} + 10x - 3 - x^{2} - 5x - 3 = 0\end{gathered}$

                            $\Large\displaystyle\begin{gathered} x^{2} + 5x -6 = 0\end{gathered}$

Calculando as raízes, temos:

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

                $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{5^{2} - 4\cdot1\cdot(-6)}}{2\cdot1}\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{25 + 24}}{2}\end{gathered} }$

                $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{49}}{2}\end{gathered} }$

                 $\Large\displaystyle\begin{gathered} = \frac{-5\pm7}{2}\end{gathered}$

Obtendo os valores das raízes:

           $\LARGE\begin{cases} x' = \frac{-5 - 7}{2} = -\frac{12}{2} = -6\\x'' = \frac{-5 + 7}{2} = \frac{2}{2} = 1\end{cases}$

Então as raízes são:

                 $\Large\displaystyle\begin{gathered} x' = -6\:\:\:e\:\:\:x'' = 1\end{gathered}$

Como queremos a soma dos valores absolutos "Sa" das raízes, então temos:

          $\Large\displaystyle\begin{gathered} S_{a} = |-6| + |1| = 6 + 1 = 7\end{gathered}$

✅ Portanto, o resultado é:

                            $\Large\displaystyle\begin{gathered} S_{a} = 7\end{gathered}$

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

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