Question

1- Resolva as inequações:
a) 5x – 9 ≤ 8x + 3
b) x² – 2x – 3< 0
c) -2x² – x + 1 ≤ 0

2- As soluções reais da inequação a seguir é o conjunto:
2x² – 5x > 2x² +3x – 8

Answer 1

$\boxed{\begin{array}{l}\rm 1)~ Resolva~as~inequac_{\!\!,}\tilde oes:\\\tt a)~\sf 5x-9\leqslant8x+3\\\tt b)~\sf x^2-2x-3<0\\\tt c~\sf -2x^2-x+1\leqslant0\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao:}\\\tt a)~\sf 5x-9\leqslant8x+3\\\sf 5x-8x\leqslant3+9\\\sf -3x\leqslant12\cdot(-1)\\\sf 3x\geqslant-12\\\sf x\geqslant-\dfrac{12}{3}\\\sf x\geqslant-4\\\tt b)~\sf x^2-2x-3<0\\\sf fac_{\!\!,}a~f(x)=x^2-2x-3\\\sf ra\acute izes~~x^2-2x-3=0\\\sf s=-\dfrac{-2}{1}=2\\\sf p=\dfrac{-3}{1}=-3\\\sf x_1=3~~e~~x_2=-1\\\sf pois~3+(-1)=2~~e~~3\cdot(-1)=-3\\\sf fazendo~o~estudo~do~sinal(veja~anexo)~temos:\\\sf f(x)>0\implies x<-1~~ou~~x>3\\\\\sf</p><p>f(x)<0\implies -2<x<3\\\sf a~soluc_{\!\!,}\tilde ao~desejada~\acute e~a~parte~negativa~portanto\\\sf S=\{x\in\mathbb{R}/-2<x<3\}\end{array}}$

$\boxed{\begin{array}{l}\tt c)~\sf-2x^2-x+1\leqslant0\\\sf fac_{\!\!,}a~~ f(x)=-2x^2-x+1\\\sf\Delta=b^2-4ac\\\sf\Delta=(-1)^2-4\cdot(-2)\cdot1\\\sf\Delta=1+8\\\sf\Delta=9\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-1)\pm\sqrt{9}}{2\cdot(-2)}\\\sf x=\dfrac{1\pm3}{-4}\begin{cases}\sf x_1=\dfrac{1+3}{4}=\dfrac{4}{4}=1\\\sf x_2=\dfrac{1-3}{4}=-\dfrac{2}{4}=-\dfrac{1}{2}\end{cases}\\\sf fazendo~o~estudo~do~sinal(veja~anexo)~temos:\\\sf f(x)\leqslant0\implies x\leqslant-1~ou~x\geqslant-\dfrac{1}{2}\\\sf S=\bigg\{x\in\mathbb{R}/x\leqslant-1~ou~x\geqslant-\dfrac{1}{2}\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\rm 2)~As~soluc_{\!\!,}\tilde oes~reais~da~inequac_{\!\!,}\tilde ao~a~seguir\\\sf \acute e~o~conjunto:\\\sf2x^2-5x>2x^2+3x-8\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao:}\\\sf2x^2-5x>2x^2+3x-8\\\sf 2x^2-2x^2-5x-3x>-8\\\sf-8x>-8\cdot(-1)\\\sf8x<8\\\sf x<\dfrac{8}{8}\\\sf x<1\\\sf S=\{x\in\mathbb{R}/x<1\} \end{array}}$