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=?

Jul 30, 2018

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

${a}^{2} + 2 b + {b}^{2} + 4 c + {c}^{2} + 6 a = - 14$

$\implies {a}^{2} + 2 \cdot a \cdot 3 + {3}^{2} + {b}^{2} + 2 \cdot b \cdot 1 + {1}^{2} + {c}^{2} + 2 \cdot c \cdot 2 + {2}^{2} = 0$

$\implies {\left(a + 3\right)}^{2} + {\left(b + 1\right)}^{2} + {\left(c + 2\right)}^{2} = 0$

So $a = - 3 , b = - 1 \mathmr{and} c = - 2$

then the value of

${a}^{2} + {b}^{2} + {c}^{2}$

$= {\left(- 3\right)}^{2} + {\left(- 1\right)}^{2} + {\left(- 2\right)}^{2} = 14$