Question

Resolva a equação
√3x+4 = x-12 , sendo U= reais

Answer 1

$\large\begin{array}{l} \mathsf{\sqrt{3x+4}=x-12\qquad\quad(i)}\\\\\\ \textsf{Mudan\c{c}a de vari\'avel:}\\\\ \mathsf{\sqrt{3x+4}=t\qquad(t\ge 0)}\\\\ \mathsf{3x+4=t^2}\\\\ \mathsf{3x=t^2-4}\\\\ \mathsf{x=\dfrac{t^2-4}{3}} \end{array}$


$\large\begin{array}{l} \textsf{Substituindo, a equa\c{c}\~ao (i) fica}\\\\ \mathsf{t=\dfrac{t^2-4}{3}-12}\\\\ \mathsf{t=\dfrac{t^2-4}{3}-\dfrac{36}{3}}\\\\ \mathsf{t=\dfrac{t^2-4-36}{3}}\\\\ \mathsf{t=\dfrac{t^2-40}{3}}\\\\ \mathsf{3t=t^2-40} \end{array}$

$\large\begin{array}{l} \mathsf{t^2-3t-40=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=1}\\\mathsf{b=-3}\\\mathsf{c=-40} \end{array} \right.\\\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-3)^2-4\cdot 1\cdot (-40)}\\\\ \mathsf{\Delta=9+160}\\\\ \mathsf{\Delta=169}\\\\ \mathsf{\Delta=13^2} \end{array}$


$\large\begin{array}{l} \mathsf{t=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{t=\dfrac{-(-3)\pm \sqrt{13^2}}{2\cdot 1}}\\\\ \mathsf{t=\dfrac{3\pm 13}{2}}\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{3+13}{2}}&~\textsf{ ou }~&\mathsf{t=\dfrac{3-13}{2}}\\\\ \mathsf{t=\dfrac{16}{2}}&~\textsf{ ou }~&\mathsf{t=\dfrac{-10}{2}}\\\\ \mathsf{t=8}&~\textsf{ ou }~&\mathsf{t=-5}\quad\textsf{(n\~ao serve, pois }\mathsf{t\ge 0}\textsf{)} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Ent\~ao ficamos com}\\\\ \mathsf{t=8}\\\\\\ \textsf{Voltando \`a vari\'avel x, temos}\\\\ \mathsf{x=\dfrac{t^2-4}{3}}\\\\ \mathsf{x=\dfrac{8^2-4}{3}}\\\\ \mathsf{x=\dfrac{64-4}{3}}\\\\ \mathsf{x=\dfrac{60}{3}}\\\\ \boxed{\begin{array}{c}\mathsf{x=20} \end{array}}\qquad\quad\checkmark \end{array}$


$\large\begin{array}{l} \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\{20\}.} \end{array}$


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


Tags: equação irracional raiz mudança de variável substituição solução resolver