Question

已知x^2+4y^2+2x-4y+2=0,求√5x^2+16y^2的值

Answer 1

2.已知x^2+4y^2+2x-4y+2=0,求√5x^2+16y^2

x^2+4y^2+2x-4y+2=0

==>(x+1)^2+(2y-1)^2=0

==>x=-1,y=1/2

===>√5x^2+16y^2=√5*(-1)^2+16*(1/2)^2=√5+4