Question

Devo colocar o x+1/x em evidência? E o que eu faço com aquele 4 que está somando?

Answer 1

$\mathbf{Primeiramente,\ desenvolveremos\ que:}\\\\ \mathrm{\bigg(x^2+\dfrac{1}{x^2}\bigg)=\bigg(x+\dfrac{1}{x}\bigg)^2-2}$

$\mathbf{Logo:}\\\\ \mathrm{\bigg(x^2+\dfrac{1}{x^2}\bigg)-3\bigg(x+\dfrac{1}{x}\bigg)+4=0\ \to}\\\\ \mathrm{\to\ \bigg(x+\dfrac{1}{x}\bigg)^2-2-3\bigg(x+\dfrac{1}{x}\bigg)+4=0\ \to}\\\\ \mathrm{\to\ \bigg(x+\dfrac{1}{x}\bigg)^2-3\bigg(x+\dfrac{1}{x}\bigg)+2=0}$

$\textbf{Substitui\c{c}\~ao:}\\\\ \boxed{\mathrm{x+\dfrac{1}{x}=u}}$

$\mathbf{Logo:}\\\\ \mathrm{u^2-3u+2=0\ \to\ a=1\ \ \|\ \ b=-3\ \ \|\ \ c=2}\\\\ \mathrm{u=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-3)\pm\sqrt{(-3)^2-4.1.2}}{2.1}=}\\\\ \mathrm{=\dfrac{3\pm\sqrt{9-8}}{2}=\dfrac{3\pm\sqrt{1}}{2}=\dfrac{3\pm1}{2}=\boxed{2}\ ou\ \boxed{1}}$

$\mathbf{Para\ u=2:}\\\\ \mathrm{x+\dfrac{1}{x}=2\ \to\ \dfrac{x^2+1}{x}=2\ \to\ x^2+1=2x\ \to}\\\\ \mathrm{x^2-2x+1=0\ \to\ a=1\ \ \|\ \ b=-2\ \ \| \ \ c=1}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-2)\pm\sqrt{(-2)^2-4.1.1}}{2.1}=}\\\\ \mathrm{=\dfrac{2\pm\sqrt{4-4}}{2}=\dfrac{2\pm0}{2}=\boxed{1}\ \to\ x\in\mathbb{R}}$

$\mathbf{Para\ u=1:}\\\\ \mathrm{x+\dfrac{1}{x}=1\ \to\ \dfrac{x^2+1}{x}=1\ \to\ x^2+1=x\ \to}\\\\ \mathrm{\to\ x^2-x+1=0\ \to\ a=1\ \ \|\ \ b=-1\ \ \|\ \ c=1}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-1)\pm\sqrt{(-1)^2-4.1.1}}{2.1}=}\\\\ \mathrm{=\dfrac{1\pm\sqrt{1-4}}{2}=\dfrac{1\pm\sqrt{-3}}{2}=\boxed{\mathrm{\dfrac{1\pm i\sqrt{3}}{2}}}\ \to\ x\in\mathbb{C}}$

$\mathbf{Portanto,\ a\ \acute{u}nica\ solu\c{c}\~ao\ real\ para\ x\ \acute{e}\ 1:}\\\\ \boxed{\boxed{\mathbf{\mathbb{S}=\{x\in\mathbb{R}\ |\ x=1\}}}}}$