Question

a^2+b^2+2c^2-4a+2c-2bc+5=0 then find the possible value for a+b-c

Answer 1

a^2 + b^2 + 2c^2 - 4a + 2c - 2bc + 5 = 0

(a - 2)^2 + (b - c)^2 + (c + 1)^2 = 0

Integer solution:

a = 2, b = -1, c = -1

a + b + c = 0

Answer 2

a²+b²+2c²-4a +2c -2bc +5=0

(a²-4a+4)+(b²-2bc+c²)+2c²+2c+5-4-c²=0

(a²-4a+4)+(b²-2bc+c²)+(c²+2c+1)=0

(a-2)²+(b-c)²+(c+1)²=0

a-2=0

a= 2

b-c=0

b=c

c+1=0

c=-1

So

a= 2, b= -1 and c= -1

so, a+b-c = 2-1+1= 2