Question

A soma das raízes na equação modular ||x-2|-7|=6, é?

Answer 1

$\large\begin{array}{l} \textsf{Primeiro vamos resolver a equa\c{c}\~ao modular e encontrar}\\\textsf{todas as suas solu\c{c}\~oes:}\\\\ \mathsf{\big||x-2|-7\big|=6}\\\\ \mathsf{|x-2|-7=\pm\,6}\\\\ \mathsf{|x-2|=\pm\,6+7}\end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{|x-2|=-6+7}&~\textsf{ ou }~&\mathsf{|x-2|=6+7}\\\\ \mathsf{|x-2|=1}&~\textsf{ ou }~&\mathsf{|x-2|=13}\\\\ \mathsf{x-2=\pm\,1}&~\textsf{ ou }~&\mathsf{x-2=\pm\,13}\\\\\mathsf{x=\pm\,1+2}&~\textsf{ ou }~&\mathsf{x=\pm\,13+2} \end{array} \end{array} \end{array}$


$\large\textsf{Por fim, temos as solu\c{c}\~oes da equa\c{c}\~ao:}$

•   $\large\begin{array}{l} \mathsf{x=-1+2} \end{array}$

$\large\begin{array}{l} \mathsf{x=1\qquad\quad\checkmark} \end{array}$


•   $\large\begin{array}{l} \mathsf{x=1+2} \end{array}$

$\large\begin{array}{l} \mathsf{x=3\qquad\quad\checkmark} \end{array}$


•   $\large\begin{array}{l} \mathsf{x=-13+2} \end{array}$

$\large\begin{array}{l} \mathsf{x=-11\qquad\quad\checkmark} \end{array}$


•   $\large\begin{array}{l} \mathsf{x=13+2} \end{array}$

$\large\begin{array}{l} \mathsf{x=15\qquad\quad\checkmark} \end{array}$


$\large\begin{array}{l} \textsf{Conjunto solu\c{c}\~ao:~~}\mathsf{S=\{-11,\,1,\,3,\,15\}.} \end{array}$


$\large\begin{array}{l} \textsf{A soma pedida \'e}\\\\\mathsf{-11+1+3+15}\\\\ =\mathsf{-10+3+15}\\\\ =\mathsf{-7+15}\\\\ =\mathsf{8}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$

Answer 2

Resposta:

8

Explicação passo-a-passo:

Tem:

||x – 2| – 7| = 6 ⇔ |x – 2| – 7 = ±6.

Logo,

|x – 2| = 13 ⇔ x – 2 = ±13

⇔ x = 15 ou x = –11

ou

|x – 2| = 1 ⇔ x – 2 = ±1

⇔ x = 3 ou x = 1.

O resultado é 15 + (–11) + 3 + 1= 8.