Question

qual a solucao da equação 2+√3x+7=x+3?

Answer 1

$2+\sqrt{3x+7}=x+3$


Faça a seguinte mudança de variável:

$\sqrt{3x+7}=t\quad(t\ge 0)\\\\ 3x+7=t^2\\\\ 3x=t^2-7\\\\ x=\dfrac{t^2-7}{3}$


Substituindo, a equação fica

$2+t=\dfrac{t^2-7}{3}+3\\\\\\ 2+t=\dfrac{t^2-7}{3}+\dfrac{9}{3}\\\\\\ 2+t=\dfrac{t^2-7+9}{3}\\\\\\ 2+t=\dfrac{t^2+2}{3}\\\\\\ 3\cdot (2+t)=t^2+2$

$6+3t=t^2+2\\\\ 0=t^2+2-6-3t\\\\ t^2-3t-4=0\quad\Rightarrow\quad\left\{ \!\begin{array}{l}a=1\\b=-3\\c=-4 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-3)^2-4\cdot 1\cdot (-4)\\\\ \Delta=9+16\\\\ \Delta=25$


$t=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ t=\dfrac{-(-3)\pm \sqrt{25}}{2\cdot 1}\\\\\\ t=\dfrac{3\pm 5}{2}\\\\\\ \begin{array}{rcl} t=\dfrac{3+5}{2}&~\text{ ou }~&t=\dfrac{3-5}{2}\\\\ t=\dfrac{8}{2}&~\text{ ou }~&t=\dfrac{-2}{2}\\\\ t=4&~\text{ ou }~&t=-1\text{ (n\~ao serve)} \end{array}$


Logo, encontramos

$t=4$


Voltando à variável $x,$

$x=\dfrac{t^2-7}{3}\\\\\\ x=\dfrac{4^2-7}{3}\\\\\\ x=\dfrac{16-7}{3}\\\\\\ x=\dfrac{9}{3}\\\\\\ x=3\qquad\checkmark$


Conjunto solução:   $S=\{3\}.$


Dúvidas? Comente.


Bons estudos! :-)