Question

Resolva em R, as desigualdades: a) |x²-x-4| ≤ 2
b) | x²-5x | >6

Answer 1

Pela propriedade do módulo dos números reais, para k, k > 0.

$\mathsf{|x| \leq k~\Longrightarrow~-k \leq x \leq k}$

Então:

$\mathsf{|x^2-x-4| \leq 2~\Longrightarrow~{-2 \leq x^2-x-4 \leq 2}}\\\\\\\begin{Bmatrix}\mathsf{x^2-x-4 \geq -2}\\\\\mathsf{e}\\\\\mathsf{x^2-x-4 \leq 2~~}\end.$

Note que a dupla desigualdade foi decomposta em duas inequações simultâneas.

Resolvendo a primeira inequação (por completamento de quadrados):

$\mathsf{x^2-x-4 \geq -2}\\\\\mathsf{x^2-x \geq 2}\\\\\mathsf{x^2-x+\frac{1}{4} \geq 2+\frac{1}{4}}\\\\\mathsf{(x-\frac{1}{2})^2 \geq \frac{8}{4}+\frac{1}{4}}\\\\\mathsf{(x-\frac{1}{2})^2 \geq \frac{9}{4}}\\\\\mathsf{|x-\frac{1}{2}| \geq \sqrt{\frac{9}{4}}}\\\\\mathsf{|x-\frac{1}{2}| \geq \frac{3}{2}~\Longrightarrow~x-\frac{1}{2} \leq -\frac{3}{2}~~ou~~x-\frac{1}{2} \geq \frac{3}{2}}\\\\\mathsf{x \leq -1~~ou~~x \geq 2~~~~(S_1)}$

Resolvendo a segunda inequação (por completamento de quadrados):

$\mathsf{x^2-x-4 \leq 2}\\\\\mathsf{x^2-x \leq 6}\\\\\mathsf{x^2-x+\frac{1}{4} \leq 6+\frac{1}{4}}\\\\\mathsf{(x-\frac{1}{2})^2 \leq \frac{24}{4}+\frac{1}{4}}\\\\\mathsf{(x-\frac{1}{2})^2 \leq \frac{25}{4}}\\\\\mathsf{|x-\frac{1}{2}| \leq \sqrt{\frac{25}{4}}}\\\\\mathsf{|x-\frac{1}{2}| \leq \frac{5}{2}~\Longrightarrow~-\frac{5}{2}\leq x-\frac{1}{2} \leq \frac{5}{2}}\\\\\mathsf{-5 \leq 2x-1 \leq 5}\\\\\mathsf{-4 \leq 2x \leq 6}\\\\\mathsf{-2 \leq x \leq 3~~~~(S_2)}$

Fazendo a intersecção das soluções, isto é: S₁ ∩ S₂.

$\fbox{ \mathsf{S_1\cap S_2=S=\begin{Bmatrix}\mathsf{x\in\mathbb{R}|~-2 \leq x \leq -1~~ou~~2 \leq x \leq 3}\end{Bmatrix}} }~~~~\mathsf{(resposta)}$

Isso tudo foi apenas a letra a, agora vamos para a letra b.

Pela propriedade do módulos dos números reais, para k, k > 0.

$\mathsf{|x|\ \textgreater \ k~\Longrightarrow~x\ \textless \ -k~~ou~~x\ \textgreater \ k}$

Então:

$\mathsf{|x^2-5x|\ \textgreater \ 6~\Longrightarrow~x^2-5x\ \textless \ -6~~ou~~x^2-5x\ \textgreater \ 6}$

Resolvendo a primeira inequação (por completamento de quadrados):

$\mathsf{x^2-5x\ \textless \ -6}\\\\\mathsf{x^2-5x+\frac{25}{4}\ \textless \ -6+\frac{25}{4}}\\\\\mathsf{(x-\frac{5}{2})^2\ \textless \ \frac{-24}{~~4}+\frac{25}{4}}\\\\\mathsf{(x-\frac{5}{2})^2\ \textless \ \frac{1}{4}}\\\\\mathsf{|x-\frac{5}{2}|\ \textless \ \sqrt{\frac{1}{4}}}\\\\\mathsf{|x-\frac{5}{2}|\ \textless \ \frac{1}{2}~\Longrightarrow~-\frac{1}{2}\ \textless \ x-\frac{5}{2}\ \textless \ \frac{1}{2}}\\\\\mathsf{-1\ \textless \ 2x-5\ \textless \ 1}\\\\\mathsf{4\ \textless \ 2x\ \textless \ 6}\\\\\mathsf{2\ \textless \ x\ \textless \ 3~~~~(S_1)}$

Resolvendo a segunda inequação (por completamento de quadrados):

$\mathsf{x^2-5x\ \textgreater \ 6}\\\\\mathsf{x^2-5x+\frac{25}{4}\ \textgreater \ 6+\frac{25}{4}}\\\\\mathsf{(x-\frac{5}{2})^2\ \textgreater \ \frac{24}{4}+\frac{25}{4}}\\\\\mathsf{(x-\frac{5}{2})^2\ \textgreater \ \frac{49}{4}}\\\\\mathsf{|x-\frac{5}{2}|\ \textgreater \ \frac{7}{2}~\Longrightarrow~x-\frac{5}{2}\ \textless \ -\frac{7}{2}~~ou~~x-\frac{5}{2}\ \textgreater \ \frac{7}{2}}\\\\\mathsf{x\ \textless \ -1~~ou~~x\ \textgreater \ 6~~~~(S_2)}$

Reunindo as soluções, isto é S₁ ∪ S₂.

$\fbox{ \mathsf{S_1\cup S_2=S=\begin{Bmatrix}\mathsf{x\in\mathbb{R}|~x\ \textless \ -1~~ou~~2\ \textless \ x\ \textless \ 3~~ou~~x\ \textgreater \ 6}\end{Bmatrix}} }~~\mathsf{(resposta)}$