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#=?

Question

5852 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-30T05:17:21

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#