Question

A terça parte de um número adicionado ao quádruplo do quadrado desse número é igual a 2 ?

Answer 1

$\dfrac{k}{3}+4k^2=2\ \to\ k+12k^2=6\ \to\ 12k^2+k-6=0\\\\ k=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-1\pm\sqrt{1^2-4(12)(-6)}}{2(12)}=\\\\ =\dfrac{-1\pm\sqrt{1+288}}{24}=\dfrac{-1\pm\sqrt{289}}{24}=\dfrac{-1\pm17}{24}\\\\ k=\dfrac{-1+17}{24}=\dfrac{16}{24}=\boxed{\dfrac{2}{3}}\ \ ou\ \ k=\dfrac{-1-17}{24}=-\dfrac{18}{24}=\boxed{-\dfrac{3}{4}}$