In a Δ ABC, 2a2+4b2 + c2=4ab+2ac,then cos B is equal to
0
1/8
3/8
7/8
2a2+4b2+c2=4ab+ac⇒a2+(2b)2−4ab+a2+c2−2ac=0⇒ (a−2b)2+(a−c)2=0⇒a=2b=ccosB=a2+c2−b22ac=c2+c2−(c2)22×c×c=2c2−c242c2⇒cosB=78