Question

Resolver a inequações:
-3x²+3x+1>0

Answer 1

Segue a resolução e a respetiva resposta

Answer 2

$\large\boxed{\begin{array}{l}\sf -3x^2+3x+1>0\\\underline{\boldsymbol{ fac_{\!\!,}a}}\\\sf f(x)=-3x^2+3x+1\\\underline{\rm zeros~de~f(x):}\\\sf -3x^2+3x+1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=3^2-4\cdot(-3)\cdot1\\\sf\Delta=9+12\\\sf\Delta=21\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-3\pm\sqrt{21}}{2\cdot(-3)}\\\\\sf x=\dfrac{-3\pm\sqrt{21}}{-6}\begin{cases}\sf x_1=\dfrac{-3+\sqrt{21}}{-6}=\dfrac{3-\sqrt{21}}{6}\\\\\sf x_2=\dfrac{-3-\sqrt{21}}{-6}=\dfrac{3+\sqrt{21}}{6} \end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\sf Resolver~a~equac_{\!\!,}\tilde ao~-3x^2+3x+1>0\\\sf significa~dizer~para~quais~valores\\\sf de~x~temos~f(x)>0.\\\sf portanto~a~resposta~\acute e\\\\\sf S=\bigg\{x\in\mathbb{R}/\dfrac{3-\sqrt{21}}{6}<x<\dfrac{3+\sqrt{21}}{6}\bigg\}\\\rm para~chegar~a~essa~conclus\tilde ao\\\rm observe~a~figura~que~eu~anexei\end{array}}$