Question

a equaçao x² + (2m - 3)x + 3 = 0 tem duas raizes reais diferentes. nessa condiçoes devemos ter. a) m _ 1/4. d) m>_ - 1/4. e) m < -2

Answer 1

Para ter duas raízes reais e diferentes devemos ter ∆>0.

$\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(2m-3)^2-4\cdot1\cdot3}\\\mathsf{\Delta=4m^2-12m+9-12}\\\mathsf{\Delta=4m^2-12m-3}\\\mathsf{4m^2-12m-3>0}$

faça

$\mathsf{f(m)=4m^2-12m-3}\\\mathsf{a=4~~b=-12~~c=-3}\\\mathsf{\Delta=b^2-4ac}\\\mathsf{\Delta=(-12)^2-4\cdot4\cdot(-3)}\\\mathsf{\Delta=144+48=192}\\\mathsf{m=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\mathsf{m=\dfrac{-(-12)\pm\sqrt{192}}{2\cdot4}}\\\mathsf{m=\dfrac{12\pm8\sqrt{3}}{8}}\\\mathsf{m=\dfrac{3\pm2\sqrt{3}}{2}}\begin{cases}\mathsf{m_{1}=\dfrac{3+2\sqrt{3}}{2}}\\\mathsf{m_{2}=\dfrac{3-2\sqrt{3}}{2}}\end{cases}$

Perceba que a parábola tem concavidade para cima pois a=4>0.fazendo o estudo do sinal temos

$\mathsf{f(m)>0\implies~m\textless\,\dfrac{3-2\sqrt{3}}{2}~~ou~~m>\dfrac{3+2\sqrt{3}}{2}}\\\mathsf{f(m)\textless0\implies\,\dfrac{3-2\sqrt{3}}{2}\textless\,m\textless\,\dfrac{3+\sqrt{3}}{2}}$

O intervalo que nos interessa é aquele que torna a função positiva, portanto

$\mathsf{m\textless\dfrac{3-2\sqrt{3}}{2}~~ou~~m>\dfrac{3+2\sqrt{3}}{2}}$

Reveja o gabarito.