Question

Resolva,nos reais a seguinte inequação : -x²+2x≤x²-5x+3<-3​

Answer 1

$-x^2+2x\leq x^2-5x+3<-3\\\\ -x^2+2x+x^2-2x\leq x^2-5x+3+x^2-2x<-3+x^2-2x\\\\ 0\leq 2x^2-7x+3<x^2-2x-3$

Primeira parte:

$2x^2-7x+3\geq 0\\\\ \mathrm{Ra\acute{i}zes}\ \to\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)}}{2(2)}=\\\\ =\dfrac{7\pm\sqrt{25}}{4}=\dfrac{7\pm5}{4}=\boxed{-\dfrac{1}{2}}\ \text{ou}\ \boxed{3}$

$\text{Para}\ f(x)\geq0\ \to\ \boxed{x\leq-\dfrac{1}{2}}\ \text{e}\ \boxed{x\geq3}$

Segunda parte:

$2x^2-7x+3<x^2-2x-3\ \therefore\ x^2-5x+6<0\\\\ \mathrm{Ra\acute{i}zes}\to\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-5)\pm\sqrt{(-5)^2-4(1)(6)}}{2(1)}=\\\\ =\dfrac{5\pm\sqrt{1}}{2}=\dfrac{5\pm1}{2}=\boxed{2}\ \text{ou}\ \boxed{3}$

$\text{Para } f(x)<0\ \to\ \boxed{2<x<3}$

Dessa forma:

$\mathbb{S}=\left\{\bigg(-\infty,-\dfrac{1}{2}\bigg]\cup[3,\infty)\right\}\cap\ ]2,3[\ =\emptyset$

$\therefore\ \boxed{S=\emptyset}$