已知a²-4a+b²-b/2+65/16=0,求a²-4根号b 的值

Question

已知a²-4a+b²-b/2+65/16=0,求a²-4根号b 的值

Answer 1

a²-4a+b²-b/2+65/16=0,
a²-4a+4+b²-b/2+1/16=0
(a-2)²+(b-1/4)²=0
a-2=0
a=2
b-1/4=0
b=1/4
a^2-4√b
=2^2-4√(1/4)
=4-4*1/2
=4-2
=2