Question

${a}^{4} + {b}^{4} = 47 \\ \\ a \times b = 1$
a + b = ???????

Answer 1

$\mathrm{ab=1\ \to\ \boxed{\mathrm{b=\dfrac{1}{a}}}}\\\\ \mathrm{a^4+b^4=47\ \to\ a^4+\dfrac{1}{a^4}=47\ \to\ \dfrac{a^8+1}{a^4}=47\ \to}\\\\ \mathrm{\to\ a^8+1=47a^4\ \to\ a^8-47a^4+1=0}\\\\ \textbf{Substitui\c{c}\~ao:}\ \boxed{\mathrm{a^4=u}}\ \to\ \mathrm{u^2-47u+1=0}\\\\ \mathrm{u=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-47)\pm\sqrt{(-47)^2-4.1.1}}{2.1}=}\\\\ \mathrm{=\dfrac{47\pm\sqrt{2209-4}}{2}=\dfrac{47\pm\sqrt{2205}}{2}=\dfrac{47\pm\sqrt{3^2.5.7^2}}{2}=\boxed{\mathrm{\dfrac{47\pm21\sqrt{5}}{2}}}}$

$\mathbf{(S_1)\ Para\ u=\dfrac{47+21\sqrt{5}}{2}:}\\\\ \mathrm{\boxed{\mathrm{a^4=\dfrac{47+21\sqrt{5}}{2}}}\ \to\ b^4=47-a^4\ \to\ \boxed{\mathrm{b^4=\dfrac{47-21\sqrt{5}}{2}}}}\\\\\\ \mathbf{(S_2)\ Para\ u=\dfrac{47-21\sqrt{5}}{2}:}\\\\ \mathrm{\boxed{\mathrm{a^4=\dfrac{47-21\sqrt{5}}{2}}}\ \to\ b^4=47-a^4\ \to\ \boxed{\mathrm{b^4=\dfrac{47+21\sqrt{5}}{2}}}}$

$\mathbf{Para\ S_1:}$

$\mathrm{a^4=\dfrac{47+21\sqrt{5}}{2}\ \to\ a^2=\pm\sqrt{\dfrac{47+21\sqrt{5}}{2}}}\\\\ \mathrm{I)\Rightarrow a^2=\sqrt{\dfrac{47+21\sqrt{5}}{2}}\ \to\ a=\pm\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}}\\\\ \mathrm{II)\Rightarrow a^2=-\sqrt{\dfrac{47+21\sqrt{5}}{2}}\ \to\ a=\pm i\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}}$

$\mathrm{b^4=\dfrac{47-21\sqrt{5}}{2}\ \to\ b^2=\pm\sqrt{\dfrac{47-21\sqrt{5}}{2}}}\\\\ \mathrm{I)\Rightarrow b^2=\sqrt{\dfrac{47-21\sqrt{5}}{2}}\ \to\ b=\pm\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}\\\\ \mathrm{II)\Rightarrow b^2=-\sqrt{\dfrac{47-21\sqrt{5}}{2}}\ \to\ b=\pm i\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}$

$\mathbf{Perceba\ que\ para\ S_2\ teremos\ as\ mesmas\ possibilidades,}\\ \mathbf{isto\ \acute{e},\ os\ resultados\ de\ a+b\ ser\~ao\ iguais\ aos\ de\ S_1.}$

$\mathbf{Portanto\, os\ possiveis\ resultados\ para\ a+b,\ con-}\\ \mathbf{siderando\ apenas\ os\ n\acute{u}meros\ reais,\ ser\~ao:}$

$\mathrm{I)\Rightarrow a=\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}};\ b=\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}\ \to}\\\\ \mathrm{\to\ \boxed{\mathbf{a+b=\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}+\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}}}$

$\mathrm{II)\Rightarrow a=\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}};\ b=-\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}\ \to}\\\\ \mathrm{\to\ \boxed{\mathbf{a+b=\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}-\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}}}$

$\mathrm{III)\Rightarrow a=-\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}};\ b=\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}\ \to}\\\\ \mathrm{\to\ \boxed{\mathbf{a+b=\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}-\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}}}}$

$\mathrm{IV)\Rightarrow a=-\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}};\ b=-\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}\ \to}\\\\ \mathrm{\to\ \boxed{\mathbf{a+b=-\bigg(\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}+\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}\bigg)}}$

$\mathbf{Respostas\ em\ \mathbb{R}:}\\\\ \mathrm{(a+b)=\bigg\{\pm\bigg(\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}+\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}}\bigg);}\\\\ \mathrm{\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}-\sqrt[4]{\dfrac{47-21\sqrt{5}}{2}};\ \sqrt[4]{\dfrac{47-21\sqrt{5}}{2}}-\sqrt[4]{\dfrac{47+21\sqrt{5}}{2}}\bigg\}}$