Respostas

2013-11-13T15:46:08-02:00
X²+8x+16=0
Δ=b²-4ac
Δ=8²-4.1.16
Δ=64-64
Δ=0
x=-b+/-raiz de Δ /2a
x=-8+/-raiz de 0 /2
x¹=-8-0/2= -4
x²=-8+0/2= -4
1 5 1
2013-11-13T15:56:22-02:00
(x+4)^2=0\\x^2+8x+16=0\\\Delta=8^2-4.1.16\\\Delta=0

Quando ▲= 0 existe uma única raiz real .

\boxed{x= - \frac{8}{2} ~~x=-4}