seja a função f(x)=2x2+(m-3)x-(m-1) determine m para que a função tenha duas raizespositivas
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Olá,, tudo bem? Sendo as duas raízes (x₁ e x₂) positivas, teremos, obrigatoriamente:
![x_{1}+x_{2}\ \textgreater \ 0\to\boxed{\dfrac{-(m-3)}{2}\ \textgreater \ 0}\,(I)\quad\text{e}\ldots\\\\x_{1}\cdot x_{2}\ \textgreater \ 0\to\boxed{\dfrac{-(m-1)}{2}\ \textgreater \ 0}\,(II) x_{1}+x_{2}\ \textgreater \ 0\to\boxed{\dfrac{-(m-3)}{2}\ \textgreater \ 0}\,(I)\quad\text{e}\ldots\\\\x_{1}\cdot x_{2}\ \textgreater \ 0\to\boxed{\dfrac{-(m-1)}{2}\ \textgreater \ 0}\,(II)](https://tex.z-dn.net/?f=x_%7B1%7D%2Bx_%7B2%7D%5C+%5Ctextgreater+%5C+0%5Cto%5Cboxed%7B%5Cdfrac%7B-%28m-3%29%7D%7B2%7D%5C+%5Ctextgreater+%5C+0%7D%5C%2C%28I%29%5Cquad%5Ctext%7Be%7D%5Cldots%5C%5C%5C%5Cx_%7B1%7D%5Ccdot+x_%7B2%7D%5C+%5Ctextgreater+%5C+0%5Cto%5Cboxed%7B%5Cdfrac%7B-%28m-1%29%7D%7B2%7D%5C+%5Ctextgreater+%5C+0%7D%5C%2C%28II%29)
![De\,\,(I):\,\, \dfrac{-(m-3)}{2}\ \textgreater \ 0\to\ldots\to\boxed{m\ \textless \ 3}\,(I)\\\\De\,\,(II):\,\, \dfrac{-(m-1)}{2}\ \textgreater \ 0\to\ldots\to\boxed{m\ \textless \ 1}\,(I) De\,\,(I):\,\, \dfrac{-(m-3)}{2}\ \textgreater \ 0\to\ldots\to\boxed{m\ \textless \ 3}\,(I)\\\\De\,\,(II):\,\, \dfrac{-(m-1)}{2}\ \textgreater \ 0\to\ldots\to\boxed{m\ \textless \ 1}\,(I)](https://tex.z-dn.net/?f=De%5C%2C%5C%2C%28I%29%3A%5C%2C%5C%2C+%5Cdfrac%7B-%28m-3%29%7D%7B2%7D%5C+%5Ctextgreater+%5C+0%5Cto%5Cldots%5Cto%5Cboxed%7Bm%5C+%5Ctextless+%5C+3%7D%5C%2C%28I%29%5C%5C%5C%5CDe%5C%2C%5C%2C%28II%29%3A%5C%2C%5C%2C+%5Cdfrac%7B-%28m-1%29%7D%7B2%7D%5C+%5Ctextgreater+%5C+0%5Cto%5Cldots%5Cto%5Cboxed%7Bm%5C+%5Ctextless+%5C+1%7D%5C%2C%28I%29)
A resposta final será a interseção de (I) e (II), ou seja:
(resposta final)
É isso!! :-)
A resposta final será a interseção de (I) e (II), ou seja:
É isso!! :-)
Anexos:
![](https://pt-static.z-dn.net/files/d31/dd23378bd0c811cfb69bb7b0b8f72174.png)
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