Question

√x²+2x+1=√7x+2 por obséquio

Answer 1

$\large\begin{array}{l} \textsf{Resolver uma equa\c{c}\~ao irracional:}\\\\ \mathsf{\sqrt{x^2+2x+1}=\sqrt{7x+2}}\\\\\\ \textsf{Eleve os dois lados da igualdade ao quadrado:}\\\\ \mathsf{\big(\sqrt{x^2+2x+1}\big)^{\!2}=\big(\sqrt{7x+2}\big)^{\!2}}\\\\ \mathsf{x^2+2x+1=7x+2}\\\\ \mathsf{x^2+2x+1-7x-2=0}\\\\ \mathsf{x^2+2x-7x+1-2=0} \end{array}$

$\large\begin{array}{l} \mathsf{x^2-5x-1=0}\quad\Rightarrow\quad\left\{\! \begin{array}{l} \mathsf{a=1}\\\mathsf{b=-5}\\\mathsf{c=-1} \end{array} \right.\\\\\\ \mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=(-5)^2-4\cdot 1\cdot (-1)}\\\\ \mathsf{\Delta=25+4}\\\\ \mathsf{\Delta=29} \end{array}$


$\large\begin{array}{l} \mathsf{x=\dfrac{-b\pm \sqrt{\Delta}}{2a}}\\\\ \mathsf{x=\dfrac{-(-5)\pm \sqrt{29}}{2\cdot 1}}\\\\ \mathsf{x=\dfrac{5\pm \sqrt{29}}{2}}\\\\ \begin{array}{rcl} \mathsf{x=\dfrac{5-\sqrt{29}}{2}}&~\textsf{ ou }~&\mathsf{x=\dfrac{5+\sqrt{29}}{2}} \end{array} \end{array}$

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$\large\begin{array}{l} \textsf{Aten\c{c}\~ao: Estamos resolvendo uma equa\c{c}\~ao irracional, ent\~ao}\\\textsf{temos que testar os valores encontrados para verificar se s\~ao}\\\textsf{solu\c{c}\~oes da equa\c{c}\~ao dada inicialmente.} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Testando }\mathsf{x_1=\frac{5-\sqrt{29}}{2}:}\\\\ \mathsf{\sqrt{x_1^2+2x_1+1}}\\\\=\mathsf{\sqrt{\bigg(\dfrac{5-\sqrt{29}}{2}\bigg)^2+\diagup\!\!\!\! 2\cdot \bigg(\dfrac{5-\sqrt{29}}{\diagup\!\!\!\! 2}\bigg)+1}}\\\\ =\mathsf{\sqrt{\dfrac{25-10\sqrt{29}+29}{4}+5-\sqrt{29}+1}}\\\\ =\mathsf{\sqrt{\dfrac{25-10\sqrt{29}+29}{4}+6-\sqrt{29}}}\\\\ =\mathsf{\sqrt{\dfrac{25-10\sqrt{29}+29}{4}+\dfrac{24-4\sqrt{29}}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{25-10\sqrt{29}+29+24-4\sqrt{29}}{4}}}\\\\=\mathsf{\sqrt{\dfrac{78-14\sqrt{29}}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{\sqrt{\dfrac{70-14\sqrt{29}+8}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{70-14\sqrt{29}}{4}+\dfrac{8}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{14\cdot (5-\sqrt{29})}{4}+2}}\\\\ =\mathsf{\sqrt{\dfrac{\diagup\!\!\!\! 2\cdot 7\cdot (5-\sqrt{29})}{\diagup\!\!\!\! 2\cdot 2}+2}}\\\\ =\mathsf{\sqrt{\dfrac{\diagup\!\!\!\! 2\cdot 7\cdot (5-\sqrt{29})}{\diagup\!\!\!\! 2\cdot 2}+2}}\\\\ =\mathsf{\sqrt{7\cdot \bigg(\dfrac{5-\sqrt{29}}{2}\bigg)+2}}\\\\ =\mathsf{\sqrt{7x_1+2}}\qquad\checkmark \end{array}$


$\large\begin{array}{l} \mathsf{x_1=\frac{5-\sqrt{29}}{2}}\textsf{ \'e solu\c{c}\~ao para a equa\c{c}\~ao dada.} \end{array}$



$\large\begin{array}{l} \bullet~~\textsf{Testando }\mathsf{x_2=\frac{5+\sqrt{29}}{2}}\textsf{ (o processo \'e an\'alogo):}\\\\ \mathsf{\sqrt{x_2^2+2x_2+1}}\\\\=\mathsf{\sqrt{\bigg(\dfrac{5+\sqrt{29}}{2}\bigg)^2+\diagup\!\!\!\! 2\cdot \bigg(\dfrac{5+\sqrt{29}}{\diagup\!\!\!\! 2}\bigg)+1}}\\\\ =\mathsf{\sqrt{\dfrac{25+10\sqrt{29}+29}{4}+5+\sqrt{29}+1}}\\\\ =\mathsf{\sqrt{\dfrac{25+10\sqrt{29}+29}{4}+6+\sqrt{29}}}\\\\ =\mathsf{\sqrt{\dfrac{25+10\sqrt{29}+29}{4}+\dfrac{24+4\sqrt{29}}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{25+10\sqrt{29}+29+24+4\sqrt{29}}{4}}}\\\\=\mathsf{\sqrt{\dfrac{78+14\sqrt{29}}{4}}} \end{array}$

$\large\begin{array}{l} =\mathsf{\sqrt{\dfrac{70+14\sqrt{29}+8}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{70+14\sqrt{29}}{4}+\dfrac{8}{4}}}\\\\ =\mathsf{\sqrt{\dfrac{14\cdot (5+\sqrt{29})}{4}+2}}\\\\ =\mathsf{\sqrt{\dfrac{\diagup\!\!\!\! 2\cdot 7\cdot (5+\sqrt{29})}{\diagup\!\!\!\! 2\cdot 2}+2}}\\\\ =\mathsf{\sqrt{\dfrac{\diagup\!\!\!\! 2\cdot 7\cdot (5+\sqrt{29})}{\diagup\!\!\!\! 2\cdot 2}+2}}\\\\ =\mathsf{\sqrt{7\cdot \bigg(\dfrac{5+\sqrt{29}}{2}\bigg)+2}}\\\\ =\mathsf{\sqrt{7x_2+2}}\qquad\checkmark \end{array}$


$\large\begin{array}{l} \mathsf{x_2=\frac{5+\sqrt{29}}{2}}\textsf{ tamb\'em \'e solu\c{c}\~ao para a equa\c{c}\~ao dada.} \end{array}$


$\large\begin{array}{l} \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\left\{\frac{5-\sqrt{29}}{2},\,\frac{5+\sqrt{29}}{2}\right\}.}\\\\\\ \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$