Question

Determine o conjunto solução para cada uma das seguintes desigualdades:
1.1) x2 + 3x + 2 ≥ 0.
1.2) 3x + 5/ x − 5 < 0.

Answer 1

1.1)

$x^2+3x+2\geq0$

$a=1,b=3,c=2\ \therefore\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\ \therefore$

$x=\dfrac{-3\pm\sqrt{3^2-4(1)(2)}}{2(1)}=\dfrac{-3\pm1}{2}=\boxed{-2}\ \text{e}\ \boxed{-1}$

$\text{Analisando o }\mathrm{gr\acute{a}fico}\ \to\ \boxed{x\leq-2}\ \text{e}\ \boxed{x\geq-1}$

$\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ |\ x\in(-\infty,-2]\cup[-1,\infty)\right\}}$

1.2)

$\dfrac{3x+5}{x-5}<0$

$\text{Caso\ 1:}\ \boxed{3x+5>0}\ \text{e}\ \boxed{x-5<0}\ \therefore$

$3x+5>0\ \therefore\ \boxed{x>-\dfrac{5}{3}}$

$x-5<0\ \therefore\ \boxed{x<5}$

$\text{Caso 2:}\ \boxed{3x+5<0}\ \text{e}\ \boxed{x-5>0}$

$3x+5<0\ \therefore\ \boxed{x<-\dfrac{5}{3}}$

$x-5>0\ \therefore\ \boxed{x>5}$

$\mathrm{H\acute{a}\ tr\hat{e}s\ intervalos\ para}\ \text{analisar:}\\\\ x<-\dfrac{5}{3}\ \to\ \text{p. ex.}\ x=-2\ \to\ \dfrac{-1}{-7}=\dfrac{1}{7}\nless0$

$-\dfrac{5}{3}<x<5\ \to\ \text{p. ex.}\ x=0\ \to\ \dfrac{5}{-5}=-1<0$

$x>5\ \to\ \text{p. ex.}\ x=6\ \to\ \dfrac{23}{1}=23\nless0$

$\mathrm{O\ \acute{u}nico\ intervalo\ para\ o\ qual\ a\ desigualdade\ \acute{e}}\ \boxed{-\dfrac{5}{3}<x<5}$

$\boxed{\mathbb{S}=\left\{x\in\mathbb{R}\ \bigg|\ -\dfrac{5}{3}<x<5\right\}}$