Question

7< x²+3 < 4x? Como resolver essa inequação

Answer 1

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Resolver a dupla desigualdade

$\mathsf{7<x^2+3<4x\qquad\quad(i)}\\\\\\ \left\{\! \begin{array}{lc} \mathsf{7<x^2+3}&\qquad\mathsf{(ii)}\\\\ \mathsf{x^2+3<4x}&\qquad\mathsf{(iii)} \end{array} \right.$


Resolvendo $\mathsf{(ii):}$

$\mathsf{7<x^2+3}\\\\ \mathsf{0<x^2+3-7}\\\\ \mathsf{x^2-4>0}\\\\ \mathsf{x^2-2^2>0}\\\\ \mathsf{x^2+2x-2x-2^2>0}\\\\ \mathsf{x(x+2)-2(x+2)>0}$

$\mathsf{(x+2)(x-2)>0}\quad\longleftarrow\quad\textsf{inequa\c{c}\~ao-produto}\qquad\mathsf{(iv)}$


Encontrando as raízes do lado esquerdo de $\mathsf{(iv):}$

$\begin{array}{rcl} \mathsf{x_1+2=0}&~\textsf{ e }~&\mathsf{x_2-2=0}\\\\ \mathsf{x_1=-2}&~\textsf{ e }~&\mathsf{x_2=2}\quad\longleftarrow\quad\textsf{(ra\'izes)} \end{array}$


Montando o quadro de sinais:

$\begin{array}{cl} \mathsf{x+2}&\qquad\mathsf{\overset{~~----}{\textsf{||||}}\!\!\!\underset{-2}{\overset{0}{\bullet}}\!\!\!\overset{++++++++}{\textsf{||||||}}\!\!\underset{2}{\bullet}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\ \mathsf{x-2}&\qquad\mathsf{\overset{~~----}{\textsf{||||}}\!\!\!\underset{-2}{\bullet}\!\!\!\overset{--------}{\textsf{||||||}}\!\!\underset{2}{\overset{0}{\bullet}}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\\\ \mathsf{(x+2)(x-2)}&\qquad\mathsf{\overset{~~++++}{\textsf{||||}}\!\!\!\underset{-2}{\overset{0}{\bullet}}\!\!\!\overset{--------}{\textsf{||||||}}\!\!\underset{2}{\overset{0}{\bullet}}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}} \end{array}$


Para que o lado esquerdo seja positivo, o intervalo de interesse é

$\mathsf{x<-2~~ou~~x>2.}$


Solução da inequação $\mathsf{(ii):}$

$\mathsf{S_{(ii)}=\left]-\infty,\,-2\right[\,\cup\,\left]2,\,+\infty\right[.}$

________


Resolvendo $\mathsf{(iii):}$

$\mathsf{x^2+3<4x}\\\\ \mathsf{x^2-4x+3<0}\\\\ \mathsf{x^2-x-3x+3<0}\\\\ \mathsf{x(x-1)-3(x-1)<0}\\\\ \mathsf{(x-1)(x-3)<0}\quad\longleftarrow\quad\textsf{inequa\c{c}\~ao-produto}\qquad\mathsf{(v)}$


Encontrando as raízes do lado esquerdo de $\mathsf{(v):}$

$\begin{array}{rcl} \mathsf{x_1-1=0}&~\textsf{ e }~&\mathsf{x_2-3=0}\\\\ \mathsf{x_1=1}&~\textsf{ e }~&\mathsf{x_2=3}\quad\longleftarrow\quad\textsf{(ra\'izes)} \end{array}$


Montando o quadro de sinais:

$\begin{array}{cl} \mathsf{x-1}&\qquad\mathsf{\overset{~~----}{\textsf{||||}}\!\!\!\underset{1}{\overset{0}{\bullet}}\!\!\!\overset{++++++++}{\textsf{||||||}}\!\!\underset{3}{\bullet}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\ \mathsf{x-3}&\qquad\mathsf{\overset{~~----}{\textsf{||||}}\!\!\!\underset{1}{\bullet}\!\!\!\overset{--------}{\textsf{||||||}}\!\!\underset{3}{\overset{0}{\bullet}}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\\\ \mathsf{(x-1)(x-3)}&\qquad\mathsf{\overset{~~++++}{\textsf{||||}}\!\!\!\underset{1}{\overset{0}{\bullet}}\!\!\!\overset{--------}{\textsf{||||||}}\!\!\underset{3}{\overset{0}{\bullet}}\!\!\overset{++++~~}{\textsf{||||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}} \end{array}$


Para que o lado esquerdo seja negativo, o intervalo de interesse é

$\mathsf{1<x<3.}$


Solução da inequação $\mathsf{(iii):}$

$\mathsf{S_{(iii)}=\left]1,\,3\right[.}$

________


A solução da dupla desigualdade $\mathsf{(i)}$ é a interseção das soluções de $\mathsf{(ii)}$ e $\mathsf{(iii)}$

$\begin{array}{cl} \mathsf{S_{(i)}}&\qquad\mathsf{\overset{~~~*****}{\textsf{|||}}\!\!\!\underset{-2}{\circ}\!\!\!\textsf{||||||}\!\!\underset{1}{\circ}\!\!\textsf{|||}\!\!\underset{2}{\circ}\!\!\overset{************~~~}{\textsf{|||}\!\!\underset{3}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\ \mathsf{S_{(ii)}}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-2}{\circ}\!\!\!\textsf{||||||}\!\!\underset{1}{\circ}\!\!\overset{**************}{\textsf{|||}\!\!\underset{2}{\bullet}\!\!\textsf{|||}}\!\!\underset{3}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}}\\\\\\ \mathsf{S_{(i)}\cap S_{(ii)}}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-2}{\circ}\!\!\!\textsf{||||||}\!\!\underset{1}{\circ}\!\!\textsf{|||}\!\!\underset{2}{\circ}\!\!\overset{*******}{\textsf{|||}}\!\!\underset{3}{\circ}\!\!\textsf{|||}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}} \end{array}$


Então o conjunto solução é

$\mathsf{S=S_{(i)}\cap S_{(ii)}}\\\\ \mathsf{S=\left]2,\,3\right[}$


ou usando a notação usual

$\mathsf{S=\{x\in\mathbb{R}:~2<x<3\}.}$


Bons estudos! :-)


Tags:  inequação dupla desigualdade quadrática álgebra