Question

Y elevado ao quadrado menos y + 4 = 0​

Answer 1

Resposta:

y^2+8y-4=0

\Delta=8^2-4\cdot1\cdot(-4)

\Delta=64+16=80

y=\dfrac{-8\pm\sqrt{80}}{2\cdot1}

Veja que, 80=2^4\cdot5, ou seja, \sqrt{80}=\sqrt{2^4\cdot5}=4\sqrt{5}

Assim, y=\dfrac{-8\pm4\sqrt{5}}{2}=-4\pm2\sqrt{5}.

y'=-4+2\sqrt{5}

y"=-4-2\sqrt{5}