Question

inequação do 2 grau x2-4x+3>0 ?

Answer 1

Olá Grazi.


Organizado e resolvendo a inequação:


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

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

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Temos uma inequação produto.

Fazendo a intersecção:


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


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


Bons estudos ! :^)


Dúvidas? comente.