Question

1 = A soma das raízes da equação :

4.2^2x-2 - 40.2^x-1 + 64 = 0 é igual a:
A) = 8
B) = 6
C) = 2
D) = - 16
E) = - 20 ​

Answer 1

$\Large\boxed{\begin{array}{l}\boldsymbol{Resposta:}~\sf6\\\boldsymbol{Explicac_{\!\!,}\tilde ao~passo~a~passo:}\\\sf 4\cdot2^{2x-2}-40\cdot2^{x-1}+64=0\\\sf 4\cdot\dfrac{(2^x)^2}{2^2}-40\cdot\dfrac{2^x}{2}+64=0\\\\\sf \backslash\!\!\!4\cdot\dfrac{(2^x)^2}{\backslash\!\!\!4}-\diagdown\!\!\!\!\!\!40\cdot\dfrac{2^x}{\diagdown\!\!\!\!2}+64=0\\\sf (2^x)^2-20\cdot2^x+64=0\\\underline{\rm fac_{\!\!,}a}\\\sf 2^x=y,y>0\end{array}}$

$\Large\boxed{\begin{array}{l}\sf y^2-20y+64=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-20)^2-4\cdot1\cdot64\\\sf\Delta=400-256\\\sf\Delta=144\\\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf y=\dfrac{-(-20)\pm\sqrt{144}}{2\cdot1}\\\\\sf y=\dfrac{20\pm12}{2}\begin{cases}\sf y_1=\dfrac{20+12}{2}=\dfrac{32}{2}=16\\\\\sf y_2=\dfrac{20-12}{2}=\dfrac{8}{2}=4\end{cases}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Quando~y=16}\\\sf 2^x=16\\\sf 2^x=2^4\\\sf x=4\\\underline{\rm Quando~y=4}\\\sf 2^x=4\\\sf 2^x=2^2\\\sf x=2\\\sf A~soma~das~ra\acute izes~\acute e~dada~por\\\sf \displaystyle\sum =4+2=6\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~B}}}}\end{array}}$