Question

As raízes da equação $x^{2} -kx+2\sqrt{2}$ =0 são números reais ''t'' e ''w'' e entre as raízes existe a relação $t^{t/w}.w^{w/t}.t^{w/t}.w^{t/w} = 512$. Os valores do parâmetros K que satisfazem essa equação são:

a)±6
b)± $4\sqrt{2}$
c)±$16\sqrt{2}$
d)±$4\sqrt[4]{2}$
e)±$16\sqrt[4]{2}$

Answer 1

Vamos começar determinando o produto das raízes "w" e "t" pela equação:

$w~.~t~=~\dfrac{c}{a}\\\\\\w~.~t~=~\dfrac{2\sqrt{2}}{1}\\\\\\\boxed{wt~=~2\sqrt{2}}\\\\\\ou\\\\\\wt~=~2~.~2^{\frac{1}{2}}\\\\\\\boxed{wt~=~2^{\frac{3}{2}}}$

Agora podemos utilizar a relação dada:

$t^{\frac{t}{w}}~.~w^{\frac{w}{t}}~.~t^{\frac{w}{t}}~.~w^{\frac{t}{w}}\\\\\\t^{\frac{t}{w}+\frac{w}{t}}~.~w^{\frac{t}{w}+\frac{w}{t}}\\\\\\(wt)^{\frac{t}{w}+\frac{w}{t}}\\\\\\(wt)^{\frac{t^2+w^2}{wt}}~=~512\\\\\\Substituindo~o ~valor~de~"wt":\\\\\\\left(2^{\frac{3}{2}}\right)^{\frac{t^2+w^2}{2^{\frac{3}{2}}}}~=~512\\\\\\\left(2\right)^{\frac{3}{2}~.~\frac{t^2+w^2}{2^{3/2}}}~=~2^9\\\\\\\frac{3}{2}~.~\frac{t^2+w^2}{2^{3/2}}~=~9\\\\\\t^2+w^2~=~9~.~\frac{2}{3}~.~2^{\frac{3}{2}}$

$t^2+w^2~=~6~.~2~\sqrt{2}\\\\\\\boxed{t^2+w^2~=~12\sqrt{2}}$

Vamos agora utilizar a equação da soma das raízes:

$w+t~=~-\frac{b}{a}\\\\\\w+t~=~-\frac{-k}{1}\\\\\\\boxed{w+t~=~k}\\\\\\Vamos~elevar~ambos~os~lados~da~equacao~ao~quadrado:\\\\\\(w+t)^2~=~k^2\\\\\\(w^2+2wt+t^2)~=~k^2\\\\\\Vamos~substituir~os~termos~conhecidos:\\\\\\(12\sqrt{2}~+~2\,.\,2\sqrt{2})~=~k^2\\\\\\k^2~=~16\sqrt{2}\\\\\\k~=~\pm\sqrt{16\sqrt{2}}\\\\\\\boxed{k~=~\pm4\sqrt[4]{2}}$