A soma das raízes da equação 2ᵡ + 2/2^ᵡ= 2√2 é?
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E aí Guilherme,
use as propriedades da potenciação que vc conhece cara:
![2^x+ \dfrac{2}{2^x}=2 \sqrt{2}\\\\
2^x+ \dfrac{2}{2^x}= \sqrt{8}\\\\
fazendo~2^x=k,~teremos:\\\\
k+ \dfrac{2}{k}= \sqrt{8}\\\\
\dfrac{k^2}{\not{k}}+ \dfrac{2}{\not{k}}= \dfrac{ \sqrt{8} k}{\not{k}}\\\\
k^2+2= \sqrt{8}k\\\\
k^2- \sqrt{8}k+2=0 2^x+ \dfrac{2}{2^x}=2 \sqrt{2}\\\\
2^x+ \dfrac{2}{2^x}= \sqrt{8}\\\\
fazendo~2^x=k,~teremos:\\\\
k+ \dfrac{2}{k}= \sqrt{8}\\\\
\dfrac{k^2}{\not{k}}+ \dfrac{2}{\not{k}}= \dfrac{ \sqrt{8} k}{\not{k}}\\\\
k^2+2= \sqrt{8}k\\\\
k^2- \sqrt{8}k+2=0](https://tex.z-dn.net/?f=2%5Ex%2B+%5Cdfrac%7B2%7D%7B2%5Ex%7D%3D2+%5Csqrt%7B2%7D%5C%5C%5C%5C%0A2%5Ex%2B+%5Cdfrac%7B2%7D%7B2%5Ex%7D%3D+%5Csqrt%7B8%7D%5C%5C%5C%5C%0Afazendo%7E2%5Ex%3Dk%2C%7Eteremos%3A%5C%5C%5C%5C%0Ak%2B+%5Cdfrac%7B2%7D%7Bk%7D%3D+%5Csqrt%7B8%7D%5C%5C%5C%5C%0A+%5Cdfrac%7Bk%5E2%7D%7B%5Cnot%7Bk%7D%7D%2B+%5Cdfrac%7B2%7D%7B%5Cnot%7Bk%7D%7D%3D++%5Cdfrac%7B+%5Csqrt%7B8%7D+k%7D%7B%5Cnot%7Bk%7D%7D%5C%5C%5C%5C%0Ak%5E2%2B2%3D+%5Csqrt%7B8%7Dk%5C%5C%5C%5C%0Ak%5E2-+%5Csqrt%7B8%7Dk%2B2%3D0++++++++++)
![\Delta=b^2-4ac\\
\Delta=(-\sqrt{8})^2-4*1*2\\
\Delta=8-8\\
\Delta=0 \Delta=b^2-4ac\\
\Delta=(-\sqrt{8})^2-4*1*2\\
\Delta=8-8\\
\Delta=0](https://tex.z-dn.net/?f=%5CDelta%3Db%5E2-4ac%5C%5C%0A%5CDelta%3D%28-%5Csqrt%7B8%7D%29%5E2-4%2A1%2A2%5C%5C%0A%5CDelta%3D8-8%5C%5C%0A%5CDelta%3D0+)
![k= \dfrac{-b\pm \sqrt{\Delta} }{2a}\\\\\\
k= \dfrac{-(- \sqrt{8})\pm \sqrt{0} }{2*1}= \dfrac{ \sqrt{8}\pm0 }{2}~\to~k'=k''= \dfrac{ \sqrt{8} }{2} k= \dfrac{-b\pm \sqrt{\Delta} }{2a}\\\\\\
k= \dfrac{-(- \sqrt{8})\pm \sqrt{0} }{2*1}= \dfrac{ \sqrt{8}\pm0 }{2}~\to~k'=k''= \dfrac{ \sqrt{8} }{2}](https://tex.z-dn.net/?f=k%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7B%5CDelta%7D+%7D%7B2a%7D%5C%5C%5C%5C%5C%5C%0Ak%3D+%5Cdfrac%7B-%28-+%5Csqrt%7B8%7D%29%5Cpm+%5Csqrt%7B0%7D++%7D%7B2%2A1%7D%3D+%5Cdfrac%7B+%5Csqrt%7B8%7D%5Cpm0+%7D%7B2%7D%7E%5Cto%7Ek%27%3Dk%27%27%3D+%5Cdfrac%7B+%5Csqrt%7B8%7D+%7D%7B2%7D++++)
Retomando a variável original, fazendo:
![2^x=k\\\\
2^x= \dfrac{ \sqrt{8} }{2}\\\\
2^x= \dfrac{2 \sqrt{2} }{2}\\\\
2^x= \sqrt{2}\\\\
\not2^x=\not2^{ \tfrac{1}{2} }\\\\
x= \dfrac{1}{2}\\\\
S=\left\{ \dfrac{1}{2}\right\} 2^x=k\\\\
2^x= \dfrac{ \sqrt{8} }{2}\\\\
2^x= \dfrac{2 \sqrt{2} }{2}\\\\
2^x= \sqrt{2}\\\\
\not2^x=\not2^{ \tfrac{1}{2} }\\\\
x= \dfrac{1}{2}\\\\
S=\left\{ \dfrac{1}{2}\right\}](https://tex.z-dn.net/?f=2%5Ex%3Dk%5C%5C%5C%5C%0A2%5Ex%3D+%5Cdfrac%7B+%5Csqrt%7B8%7D+%7D%7B2%7D%5C%5C%5C%5C%0A2%5Ex%3D++%5Cdfrac%7B2+%5Csqrt%7B2%7D+%7D%7B2%7D%5C%5C%5C%5C%0A2%5Ex%3D+%5Csqrt%7B2%7D%5C%5C%5C%5C%0A%5Cnot2%5Ex%3D%5Cnot2%5E%7B+%5Ctfrac%7B1%7D%7B2%7D+%7D%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%0AS%3D%5Cleft%5C%7B+%5Cdfrac%7B1%7D%7B2%7D%5Cright%5C%7D+++++)
OBS.:Esta equação possui apenas uma raiz.
Espero ter ajudado e tenha ótimos estudos =))
use as propriedades da potenciação que vc conhece cara:
Retomando a variável original, fazendo:
OBS.:Esta equação possui apenas uma raiz.
Espero ter ajudado e tenha ótimos estudos =))
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