Resolva a inequação abaixo:
x² – 4x +3 ≥ 0
Soluções para a tarefa
Respondido por
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^2-4x+3\ \geq 0}\\\\\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-4)^2-4\cdot1\cdot3}\\\mathsf{\Delta=16-12}\\\mathsf{\Delta=4}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%5E2-4x%2B3%5C+%5Cgeq+0%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7B%5CDelta%3Db%5E2-4ac%7D%5C%5C%5Cmathsf%7B%5CDelta%3D%28-4%29%5E2-4%5Ccdot1%5Ccdot3%7D%5C%5C%5Cmathsf%7B%5CDelta%3D16-12%7D%5C%5C%5Cmathsf%7B%5CDelta%3D4%7D)
![\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=\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}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx%3D%5Cdfrac%7B-b%5Cpm%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%5E%2B%3D%5Cdfrac%7B-%28-4%29%2B%5Csqrt%7B4%7D%7D%7B2%5Ccdot1%7D%5Cqquad%5Cqquad%5Cqquad%5Cqquad+x%5E-%3D%5Cdfrac%7B%28-4%29-%5Csqrt%7B4%7D%7D%7B2%5Ccdot1%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%5E%2B%3D%5Cdfrac%7B4%2B2%7D%7B2%7D%5Cqquad%5Cqquad%5Cqquad%5Cqquad%5Cqquad%7E%7E%7Ex%5E-%3D%5Cdfrac%7B4-2%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx%5E%2B%3D%5Cdfrac%7B6%7D%7B2%7D%5Cqquad%5Cqquad%5Cqquad%5Cqquad%5Cqquad%5Cqquad+%7E%7Ex%5E-%3D%5Cdfrac%7B2%7D%7B2%7D%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cmathsf%7Bx%5E%2B%3D3%7D%7D%5Cqquad%5Cqquad%5Cqquad%5Cqquad%5Cqquad%5Cqquad+%5Cboxed%7B%5Cmathsf%7Bx%5E-%3D1%7D%7D%5C%5C%5C%5C%5C%5C%5C%5C%5Cmathsf%7B%28x-3%29%5Ccdot%28x-1%29%3Dx%5E2-4x%2B3%7D%5C%5C%5C%5C%5C%5C%5Cmathsf%7Bx-3%7E%5CRightarrow%7Ex%3D3%7D)
![\mathsf{x-1~\Rightarrow~x=1} \mathsf{x-1~\Rightarrow~x=1}](https://tex.z-dn.net/?f=%5Cmathsf%7Bx-1%7E%5CRightarrow%7Ex%3D1%7D)
__________________________
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} \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}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cmathsf%7B%28x-1%29%5Cqquad%5Cqquad%5Cqquad+%7B_%5Cunderline%7B%7E--%7D%7D+%7B_%5Cunderline%7B----%7D%7D%5Cunderset1%5Cbullet%7D%7D%7B_%5Cunderline%7B%7E%2B%2B%2B%2B%2B%2B%2B%2B%2B%2B%2B%2B%2B%7D%7D_%5Cblacktriangleright%7D%7D%7D%7D%7D%5C%5C%5C%5C%5Cmathsf%7B%28x-3%29%5Cqquad%5Cqquad%5Cqquad+%7B_%5Cunderline%7B%7E-------------%7D%7D%7D%5Cunderset3%5Cbullet%7B_%5Cunderline%7B%2B%2B%2B%2B%2B%2B%7D%7D_%5Cblacktriangleright%7D%7D%7D%5C%5C%5C%5C%5Cmathsf%7B%28x-1%29%5Ccdot%28x-3%29%5Cqquad+%7B_%5Cunderline%7B%7E%2B%2B%2B%2B%2B%2B%7D%7D%5Cunderset1%5Cbullet%7B_%5Cunderline%7B------%7E%7D%7D%5Cunderset3%5Cbullet%7B_%5Cunderline%7B%2B%2B%2B%2B%2B%7D%7D_%5Cblacktriangleright%7D%7D%7D%5Cend%7Bcases%7D)
![\mathsf{(x-1)\cdot(x-3)\ \geq 0}\\\\\\\boxed{\boxed{\mathsf{S:\{x\in\mathbb{R}: x\leq 1~ou~x\geq 3\}}}} \mathsf{(x-1)\cdot(x-3)\ \geq 0}\\\\\\\boxed{\boxed{\mathsf{S:\{x\in\mathbb{R}: x\leq 1~ou~x\geq 3\}}}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%28x-1%29%5Ccdot%28x-3%29%5C+%5Cgeq+0%7D%5C%5C%5C%5C%5C%5C%5Cboxed%7B%5Cboxed%7B%5Cmathsf%7BS%3A%5C%7Bx%5Cin%5Cmathbb%7BR%7D%3A+x%5Cleq+1%7Eou%7Ex%5Cgeq+3%5C%7D%7D%7D%7D)
Bons estudos ! :^)
Dúvidas? comente.
Organizado e resolvendo a inequação:
__________________________
Temos uma inequação produto.
Fazendo a intersecção:
Bons estudos ! :^)
Dúvidas? comente.
Respondido por
5
Boa tarde
X(x-4)=0
X1=0
X-4
X2=4
Espero ter ajudado
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