Question

Como deixar em forma de bascara
5 sobre x ao quadrado + 1 sobre x - 6 = 0

Answer 1

$\dfrac{5}{x^2}~ +~\dfrac{1}{x}~ -~6=0~\to~~mmc=x^2\\\\\\ \dfrac{5+x-6x^2}{\not x^2} = \dfrac{0}{\not x^2}~ \to \\\\ \\-6 x^{2} +x+5=0\\\\ a=-6;b=1;c=5\\\\\\ \Delta=b^2-4ac\to \Delta=1^2-4.(-6).5\to \Delta=1+120\to \boxed{\Delta=121} \\\\ x' \neq x''$


$\\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \to x= \dfrac{-1\pm \sqrt{121} }{2.(-6)} \to x= \dfrac{-1\pm 11 }{-12} \to \\\\ \\ x'= \dfrac{-1+ 11 }{-12} \to x'= \dfrac{ 10 }{-12} \to \boxed{x'= -\dfrac{5 }{6}} \\\\\\ x''= \dfrac{-1- 11 }{-12} \to x''= \dfrac{-12 }{-12} \to \boxed{x''=1}\\\\\\\\ \boxed{\boxed{S=\{-\frac{5 }{6};1 \}}}$