Question

if a, b and c are real numbers:
a^2+2b=7, b^2+4c=-7, c^2+6a=-14
find a^2 +b^2 +c^2

Answer 1

Add all the three equations. The LHS now becomes (after rearranging a bit):

a^2 + 6a + b^2 + 2b + c^2 + 4c
=> (a+3)^2 - 9 + (b+1)^2 - 1 + (c+2)^2 - 4
=> (a+3)^2 + (b+1)^2 + (c+2)^2 - 14

The RHS is also -14 since (6 - 7 - 13) = -14.

The -14 term cancels out on both ends leaving us with:

(a+3)^2 + (b+1)^2 + (c+2)^2 = 0
which implies, a = -3, b = -1 and c = -2

Now, a^2 + b^2 + c^2 = (-3)^2 + (-1)^2 + (-2)^2 = 9+1+4 = 14

Answer 2

Hii

Plz refer to the attachment for the answer..