Question

$Sabendo~~que~~k^4^x+k^-^4^x=34,~~calcule~~o~~valor~~de~~k^x-k^-^x.$

Answer 1

$\Large\boxed{\begin{array}{l}\sf k^{4x}+k^{-4x}=34,k^x-k^{-x}=?.\\\sf Partindo\,da\,primeira \,express\tilde ao\\\sf (k^x-k^{-x})^2=k^{2x}-2k^x\cdot k^{-x}+k^{-2x}\\\sf chamando\, k^x-k^{-x}=y,y>0\\\sf y^2=k^{2x}-2+k^{2x}\\\sf y^2+2=k^{2x}+k^{-2x}\\\sf elevando\,ambos\,os\,membros\\\sf (y^2+2)^2=(k^{2x}+k^{-2x})^2\\\sf y^4+4y^2+4=k^{4x}+2+k^{-4x}\\\sf y^4+4y^2+4=k^{4x}+k^{-4x}+2\\\sf substituindo\,temos:\\\sf y^4+4y^2+4=34+2\\\sf y^4+4y^2+4=36\\\sf y^4+4y^2+4-36=0\\\sf y^4+4y^2-32=0\end{array}}$

$\Large\boxed{\begin{array}{l}\sf (y^2)^2+4y^2-32=0\\\sf fac_{\!\!,}a~y^2=w,com\,w>0.\\\sf w^2+4w-32=0\\\sf\Delta=b^2-4ac\\\sf\Delta=4^2-4\cdot1\cdot(-32)\\\sf\Delta=16+128\\\sf\Delta=144\\\sf w=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf w=\dfrac{-4\pm\sqrt{144}}{2\cdot1}\\\\\sf w=\dfrac{-4\pm12}{2}\begin{cases}\sf w_1=\dfrac{-4+12}{2}=\dfrac{8}{2}=4\\\\\sf w_2=\dfrac{-4-12}{2}=-\dfrac{16}{2}=-8\end{cases}\\\\\sf como~w>0,~a\,resposta\,\acute e~w=4.\end{array}}$

$\Large\boxed{\begin{array}{l}\sf substituindo\,temos:\\\sf y^2=4\\\sf y=\pm\sqrt{4}\\\sf y=\pm2\\\sf como~deve-se~ter~y>0\\\sf considera-se~y=2~como~resposta\\\sf portanto\\\huge\boxed{\boxed{\boxed{\boxed{\sf k^x-k^{-x}=2}}}}\end{array}}$