Question

resposta da equacao irracional √7x-3-1=x

Answer 1

$\sqrt{7x-3}-1=x\\\sqrt{7x-3}=x+1\\(\sqrt{7x-3})^2=(x+1)^2\\7x-3=x^2+2x+1\\x^2+2x+1-7x+3=0\\x^2-5x+4=0\\\Delta=25-16\\\Delta=9\\\ x=\frac{5\pm3}{2}\\ x=4\ ou\ x=1$

Testemos:

$\sqrt{7(4)-3}-1=(4)\\\sqrt{28-3}-1=4\\\sqrt{25}-1=4\\5-1=4\\4=4\ (v)$

$\sqrt{7(1)-3}-1=(1)\\\sqrt{7-3}-1=1\\\sqrt4-1=1\\2-1=1\\1=1\ (v)$

Answer 2

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

                   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \{1,\,4\}\:\:\:}}\end{gathered} }$

Resolvendo equação irracional:

                  $\Large\displaystyle\begin{gathered} \sqrt{7x - 3} - 1 = x\end{gathered}$

                           $\Large\displaystyle\begin{gathered} \sqrt{7x - 3} = x + 1\end{gathered}$

                     $\Large\displaystyle\begin{gathered} (\sqrt[\!\diagup\!]{7x - 3})^{\!\diagup\!\!\!\!2} = (x + 1)^{2}\end{gathered}$

                               $\Large\displaystyle\begin{gathered} 7x - 3 = x^{2} + 2x + 1\end{gathered}$

 $\Large\displaystyle\begin{gathered} 7x - 3 - x^{2} - 2x - 1 = 0\end{gathered}$

                 $\Large\displaystyle\begin{gathered} -x^{2} + 5x - 4 = 0\end{gathered}$

Chegamos à equação do segundo grau, cujos coeficientes são:

                          $\Large\begin{cases} a = -1\\b = 5\\c = -4\end{cases}$

Aplicando a fórmula resolutiva da equação do segundo grau, temos::

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

                 $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{5^{2} - 4\cdot(-1)\cdot(-4)}}{2\cdot(-1)}\end{gathered} }$

                 $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{25 - 16}}{-2}\end{gathered} }$

                 $\Large\displaystyle\text{ \begin{gathered} = \frac{-5\pm\sqrt{9}}{-2}\end{gathered} }$

                 $\Large\displaystyle\begin{gathered} = \frac{-5\pm3}{-2}\end{gathered}$

Então, chegamos as seguintes raízes:

      $\LARGE\begin{cases} x' = \frac{-5+ 3}{-2} = \frac{-2}{-2} = 1\\x'' = \frac{-5 - 3}{-2} = \frac{-8}{-2} = 4\end{cases}$

Testando as raízes temos:

• x' = 1

         $\Large\displaystyle\begin{gathered} \sqrt{7\cdot1 - 3} - 1 = 1\end{gathered}$

                $\Large\displaystyle\begin{gathered} \sqrt{7 - 3} - 1 = 1\end{gathered}$

                         $\Large\displaystyle\begin{gathered} \sqrt{4} - 1 = 1\end{gathered}$

                             $\Large\displaystyle\begin{gathered} 2 - 1 = 1\end{gathered}$

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

• x'' = 4

          $\Large\displaystyle\begin{gathered} \sqrt{7\cdot4 - 3} - 1 = 4\end{gathered}$

             $\Large\displaystyle\begin{gathered} \sqrt{28 - 3} - 1 = 4\end{gathered}$

                      $\Large\displaystyle\begin{gathered} \sqrt{25} - 1 = 4\end{gathered}$

                             $\Large\displaystyle\begin{gathered} 5 - 1 = 4\end{gathered}$

                                      $\Large\displaystyle\begin{gathered} 4 = 4\end{gathered}$

✅ Portanto, o conjunto solução é:

                                $\Large\displaystyle\begin{gathered} S = \{1,\,4\}\end{gathered}$

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

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