• Matéria: Matemática
  • Autor: Mariana0701
  • Perguntado 9 anos atrás

Resolva a inequação abaixo:
x² – 4x +3 ≥ 0

Respostas

respondido por: superaks
15
Olá Mariana.


Organizado e resolvendo a inequação:


\mathsf{x^2-4x+3\ \geq 0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-4)^2-4\cdot1\cdot3}\\\mathsf{\Delta=16-12}\\\mathsf{\Delta=4}

\mathsf{x=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\\mathsf{x^+=\dfrac{-(-4)+\sqrt{4}}{2\cdot1}\qquad\qquad\qquad\qquad x^-=\dfrac{(-4)-\sqrt{4}}{2\cdot1}}\\\\\\\mathsf{x^+=\dfrac{4+2}{2}\qquad\qquad\qquad\qquad\qquad~~~x^-=\dfrac{4-2}{2}}\\\\\\\mathsf{x^+=\dfrac{6}{2}\qquad\qquad\qquad\qquad\qquad\qquad ~~x^-=\dfrac{2}{2}}\\\\\\\boxed{\mathsf{x^+=3}}\qquad\qquad\qquad\qquad\qquad\qquad \boxed{\mathsf{x^-=1}}\\\\\\\\\mathsf{(x-3)\cdot(x-1)=x^2-4x+3}\\\\\\\mathsf{x-3~\Rightarrow~x=3}
\mathsf{x-1~\Rightarrow~x=1}

__________________________

Temos uma inequação produto. 

Fazendo a intersecção:


\begin{cases}\mathsf{(x-1)\qquad\qquad\qquad {_\underline{~--}} {_\underline{----}}\underset1\bullet}}{_\underline{~+++++++++++++}}_\blacktriangleright}}}}}\\\\\mathsf{(x-3)\qquad\qquad\qquad {_\underline{~-------------}}}\underset3\bullet{_\underline{++++++}}_\blacktriangleright}}}\\\\\mathsf{(x-1)\cdot(x-3)\qquad {_\underline{~++++++}}\underset1\bullet{_\underline{------~}}\underset3\bullet{_\underline{+++++}}_\blacktriangleright}}}\end{cases}


\mathsf{(x-1)\cdot(x-3)\ \geq 0}\\\\\\\boxed{\boxed{\mathsf{S:\{x\in\mathbb{R}: x\leq 1~ou~x\geq 3\}}}}


Bons estudos ! :^)


Dúvidas? comente.
respondido por: lelepalominope9z42
5

Boa tarde

X(x-4)=0

X1=0

X-4

X2=4

Espero ter ajudado



Perguntas similares