If a,b,c are real numbers such that a^2+2b=7, b^2+4c=-7,c^2+6a=-14 then the value of a^2+b^2+c^2=?

1 Answer
Jul 30, 2018

If a,b,c are real numbers such that a^2+2b=7, b^2+4c=-7,c^2+6a=-14

Adding we get

a^2+2b+b^2+4c+c^2+6a=-14

=>a^2+2*a*3+3^2+b^2+2*b*1+1^2+c^2+2*c*2+2^2=0

=>(a+3)^2+(b+1)^2+(c+2)^2=0

So a=-3,b=-1andc=-2

then the value of

a^2+b^2+c^2

=(-3)^2+(-1)^2+(-2)^2=14