Question

1)Calcule as equações:a)$2 x^{2} +8x+12=0$ b)$x^{2} +6x-8=0$

Answer 1

$x = \frac{-b \pm \sqrt{b^2 -4*a*c}}{2*a} \\ \\ a)\\2x^2 + 8x + 12 = 0$

a = 2, b = 8, c = 12
Δ=b2−4ac
Δ=(8)2−4*(2)*(12)
Δ=64−96
Δ=−32
Não existe solução para os números reais (R):

Solução para o conjuntos dos números complexos (C):

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-8 \pm \sqrt{-32}}{2*2} \\ \\ x = \frac{-8 \pm 4\sqrt{2}i}{4} \\ \\ x' = \frac{-8 + 4\sqrt{2}i}{4} \\ \\ x' = \frac{-2 + \sqrt{2}i} \\ \\ \\ x'' = \frac{-8 - 4\sqrt{2}i}{4} \\ \\ x'' = \frac{-2 - \sqrt{2}i}$

S= {$\frac{-2 + \sqrt{2}i}, \frac{-2 - \sqrt{2}i}$}

==========================================
$b)\\x^2 + 6x - 8 = 0$
a=1, b=6, c=−8
Δ=b2−4ac
Δ=(6)2−4*(1)*(−8)
Δ=36+32
Δ=68

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-6 \pm \sqrt{68}}{2*1} \\ \\ x = \frac{-6 \pm 2\sqrt{17}}{2} \\ \\ x' = \frac{-6 + 2\sqrt{17}}{2} \\ \\ x' = \frac{-3 +\sqrt{17}} \\ \\ \\ x'' = \frac{-6 -2\sqrt{17}}{2} \\ \\ x'' = \frac{-3 -\sqrt{17}}$

S = {$\frac{-3+\sqrt{17}}, \frac{-3 -\sqrt{17}}$}