Question

4- Resolva as inequações abaixo:
a) x2 – 9x + 10 > 0.

Answer 1

$\boxed{\begin{array}{l}\sf x^2-9x+10>0\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf f(x)=x^2-9x+10\\\sf devemos~dizer~para~quais~valores~de~x\\\sf ocorrer~f(x)>0.\\\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Ra\acute izes~de~f(x):}\\\sf x^2-9x+10=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-9)^2-4\cdot1\cdot10\\\sf\Delta=81-40\\\sf\Delta=41\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-9)\pm\sqrt{41}}{2\cdot1}\\\\\sf x=\dfrac{9\pm\sqrt{41}}{2}\begin{cases}\sf x_1=\dfrac{9+\sqrt{41}}{2}\\\\\sf x_2=\dfrac{9-\sqrt{41}}{2}\end{cases}\\\\\sf f(x)>0\Longleftrightarrow x<\dfrac{9-\sqrt{41}}{2}~ou~x>\dfrac{9+\sqrt{41}}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\sf A~soluc_{\!\!,}\tilde ao~da~inequac_{\!\!,}\tilde ao~\acute e\\\sf S=\bigg\{x\in\mathbb{R}/ x<\dfrac{9-\sqrt{41}}{2}~ou~x>\dfrac{9+\sqrt{41}}{2}\bigg\}\end{array}}$