Qual o valor da soluçãox^(1/2) - x^(1/4) = 2
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Boa noite Camanda !
Solução!
Uma questão de álgebra bem interessante.
Para resolve-la vamos ter que usar alguns artifícios,para sairmos em uma equação do segundo grau.
Veja que vou usar o expoente igual a 1/4.
![[ x^{ \frac{1}{4}} ]^{2}- x^{ \frac{1}{2} }=2\\\\\
y=[ x^{ \frac{1}{4}} ]^{2}](https://tex.z-dn.net/?f=%5B+x%5E%7B+%5Cfrac%7B1%7D%7B4%7D%7D+%5D%5E%7B2%7D-+x%5E%7B+%5Cfrac%7B1%7D%7B2%7D+%7D%3D2%5C%5C%5C%5C%5C%0Ay%3D%5B+x%5E%7B+%5Cfrac%7B1%7D%7B4%7D%7D+%5D%5E%7B2%7D+++ "[ x^{ \frac{1}{4}} ]^{2}- x^{ \frac{1}{2} }=2\\\\\
y=[ x^{ \frac{1}{4}} ]^{2}")
![y^{2}-y=2\\\\
y^{2}-y-2=0\\\\\\\\\
y= \dfrac{1\pm \sqrt{(-1)^{2}-4.1.2 } }{2.1}\\\\\\
y= \dfrac{1\pm \sqrt{9 } }{2}\\\\\\
y= \dfrac{1\pm 3 }{2}\\\\\\\\\\\
y_{1} = \dfrac{1+3}{2}= \dfrac{4}{2}=2\\\\\\
y_{2} = \dfrac{1-3}{2}= \dfrac{-2}{2}=-1\Leftarrow ~~nao ~~serve\\\\\\](https://tex.z-dn.net/?f=y%5E%7B2%7D-y%3D2%5C%5C%5C%5C%0Ay%5E%7B2%7D-y-2%3D0%5C%5C%5C%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B1%5Cpm+%5Csqrt%7B%28-1%29%5E%7B2%7D-4.1.2+%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B1%5Cpm+%5Csqrt%7B9+%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B1%5Cpm+3+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+y_%7B1%7D+%3D+%5Cdfrac%7B1%2B3%7D%7B2%7D%3D+%5Cdfrac%7B4%7D%7B2%7D%3D2%5C%5C%5C%5C%5C%5C%0Ay_%7B2%7D+%3D+%5Cdfrac%7B1-3%7D%7B2%7D%3D+%5Cdfrac%7B-2%7D%7B2%7D%3D-1%5CLeftarrow+%7E%7Enao+%7E%7Eserve%5C%5C%5C%5C%5C%5C%0A++%0A++++%0A++ "y^{2}-y=2\\\\
y^{2}-y-2=0\\\\\\\\\
y= \dfrac{1\pm \sqrt{(-1)^{2}-4.1.2 } }{2.1}\\\\\\
y= \dfrac{1\pm \sqrt{9 } }{2}\\\\\\
y= \dfrac{1\pm 3 }{2}\\\\\\\\\\\
y_{1} = \dfrac{1+3}{2}= \dfrac{4}{2}=2\\\\\\
y_{2} = \dfrac{1-3}{2}= \dfrac{-2}{2}=-1\Leftarrow ~~nao ~~serve\\\\\\")
Fazendo!
![x^{1/4}=y_{1} \\\\\\
x^{1/4}=2 \\\\
\sqrt[4]{x} =2 \\\\
(\sqrt[4]{x})^{4} =(2)^{4} \\\\
x=16](https://tex.z-dn.net/?f=+x%5E%7B1%2F4%7D%3Dy_%7B1%7D+%5C%5C%5C%5C%5C%5C%0A++x%5E%7B1%2F4%7D%3D2+%5C%5C%5C%5C%0A+++%5Csqrt%5B4%5D%7Bx%7D+%3D2+%5C%5C%5C%5C+%0A%28%5Csqrt%5B4%5D%7Bx%7D%29%5E%7B4%7D++%3D%282%29%5E%7B4%7D++%5C%5C%5C%5C%0Ax%3D16+ "x^{1/4}=y_{1} \\\\\\
x^{1/4}=2 \\\\
\sqrt[4]{x} =2 \\\\
(\sqrt[4]{x})^{4} =(2)^{4} \\\\
x=16")
Boa noite!
Bons estudos!
Solução!
Uma questão de álgebra bem interessante.
Para resolve-la vamos ter que usar alguns artifícios,para sairmos em uma equação do segundo grau.
Veja que vou usar o expoente igual a 1/4.
Fazendo!
Boa noite!
Bons estudos!
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