Question

Resolver a inequação 2^2x - 6.2^x + 8 ≤0

Answer 1

Olá Joana.


Organizando e resolvendo a equação:


$\mathsf{2^{2x}-6\cdot 2^{x}+8 \leq 0}\\\\\mathsf{\underbrace{\mathsf{2}}^x\cdot2^x-6\cdot 2^x+8 \leq 0}\\\mathsf{ ~~y}\\\\\mathsf{y^2-6y+8 \leq 0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-6)^2-4\cdot1\cdot8}\\\mathsf{\Delta=36-32}\\\mathsf{\Delta=4}\\\\\\\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}$

$\mathsf{y^+=\dfrac{-(-6)+\sqrt{4}}{2\cdot1}\qquad\qquad\qquad\qquad y^-=\dfrac{-(-6)-\sqrt{4}}{2\cdot 1}}\\\\\\\mathsf{y^+=\dfrac{6+2}{2}\qquad\qquad\qquad\qquad\qquad~~~ y^-=\dfrac{6-2}{2}}\\\\\\\mathsf{y^+=\dfrac{8}{2}\qquad\qquad\qquad\qquad\qquad\qquad~~ y^-=\dfrac{4}{2}}\\\\\\\boxed{\mathsf{y^+=4}}\qquad\qquad\qquad\qquad\qquad\qquad \boxed{\mathsf{y^-=2}}\\\\\\\\\mathsf{(y-4)\cdot(y-2)=y^2-6y+8}\\\\\mathsf{(y-4)\cdot(y-2) \leq 0\qquad\gets\qquad Inequa\c{c}\~ao~produto}$


$\mathsf{y-4=0~\Rightarrow~y=4}\\\\\mathsf{y-2=0~\Rightarrow~y=2}$



$\begin{cases}\mathsf{(y-4)\qquad\qquad\qquad\qquad _{\underline{~-----~}}\underset2\bullet{_\underline{~-----~}}\underset4\bullet {_\underline{~+++++~}}_\blacktriangleright}}}\\\\\\\mathsf{(y-2)\qquad\qquad\qquad\qquad _{\underline{~-----~}}\underset2\bullet{_\underline{~+++++~}}\underset4\bullet {_\underline{~+++++~}}_\blacktriangleright}}}\\\\\\\mathsf{(y-4)\cdot(y-2)\qquad \qquad~_\underline{~+++++~}}~\underset2\bullet{_\underline{~-----~}}\underset4\bullet{_\underline{~+++++~}\blacktriangleright}}\end$


$\mathsf{2 \leq y \leq 4}\\\\\mathsf{2 \leq 2^x \leq 4}\\\\\mathsf{2^1 \leq 2^x \leq 2^2}\\\\\mathsf{1 \leq x \leq 2}\\\\\\\\\boxed{\boxed{\mathsf{S:\{x\in\mathbb{R}: 1 \leq x \leq 2\}}}}$


Dúvidas? comente.