Question

y ao quadrado+8y-4=0 me respondam por favor!

Answer 1

$y^2+8y-4=0$

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

$\Delta=64+16=80$

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

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

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

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

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

Answer 2

y²+8y-4=0
a=1, b=8, c=-4
Δ=b²-4.a.c
Δ=b²-4.a.c
Δ=8²-4.1.(-4)
Δ=64+16
Δ=80

80 | 2
40 | 2
20 | 2
10 | 2
  5 | 5
  1
√80 = √2².√2².√5 = 2.2.√5 = 4√5

y=(-b+-√Δ)/2.a
y=(-8+-√80)/2.1
y=(-8+-4√5)/2
y'=(-8+4√5)/2 = 2(-4+2√5)/2 = -4+2√5
y"=(-8-4√5)/2 = 2(-4-2√5)/2 = -4-2√5

Solução: {-4-2√5; -4+2√5}