Question

resolva a inequação simultanea 8<x2-2x-8<0​

Answer 1

$\large\boxed{\begin{array}{l}\rm 8<x^2-2x-8<0\implies\begin{cases}\rm x^2-2x-8>8~(I)\\\rm x^2-2x-8<0~~(II)\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm x^2-2x-8>8\\\rm x^2-2x-8-8>0\\\rm x^2-2x-16>0\\\underline{\sf fac_{\!\!,}a}\\\rm f(x)=x^2-2x-16\\\underline{\sf zeros~de~f(x):}\\\rm x^2-2x-16=0\\\rm x^2-2x+1-17=0\\\rm (x-1)^2=17\\\rm x-1=\pm\sqrt{17}\\\rm x-1=\sqrt{17}\longrightarrow x= 1+\sqrt{17}\\\rm x-1=-\sqrt{17}\longrightarrow x=1-\sqrt{17}\\\rm f(x)>0\iff x<1-\sqrt{17}~ou~x>1+\sqrt{17}\\\rm S_1=\{x\in\mathbb{R}/x<1-\sqrt{17}~ou~x>1+\sqrt{17}\}\end{array}}$

$\large\boxed{\begin{array}{l}\rm x^2-2x-8<0\\\underline{\sf fac_{\!\!,}a}\\\rm g(x)=x^2-2x-8\\\underline{\sf zeros~de~g(x):}\\\rm x^2-2x-8=0\\\rm x^2-2x+1-9=0\\\rm (x-1)^2=9\\\rm x-1=\pm\sqrt{9}\\\rm x-1=\pm3\\\rm x-1=3\longrightarrow x=3+1=4\\\rm x-1=-3\longrightarrow x=1-3=-2\\\rm g(x)<0\iff -2<x<4\\\rm S_2=\{x\in\mathbb{R}/-2<x<4\}\end{array}}$

$\large\boxed{\begin{array}{l}\rm S_1\cap S_2=\varnothing\\\rm S=\varnothing\end{array}}$