Question

If $a^2+2b=7,b^2+4c=-7and{c}^2+6a=-14$ then the value of $(a^2+b^2+c^2)$ is

• A. $14$
• B. $25$
• C. $36$
• D. $47$

Answer 1

Answer: (A)

$14$

Consider the given equations $a^2+2b=7$, $b^2+4c=-7$ and $c^2+6a=-14$ and add them as follows:

$a^2+2b+b^2+4c+c^2+6a=7-7-14\Rightarrow a^2+b^2+c^2+6a+2b+4c=-14\Rightarrow a^2+b^2+c^2+6a+2b+4c+14=0\Rightarrow (a^2+6a+9)+(b^2+2b+1)+(c^2+4c+4)=0\Rightarrow (a+3{)}^2+(b+1{)}^2+(c+2{)}^2=0(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow (a+3{)}^2=0,(b+1{)}^2=0,(c+2{)}^2=0\Rightarrow a+3=0,b+1=0,c+2=0\Rightarrow a=-3,b=-1,c=-2\Rightarrow a^2=(-3{)}^2,b^2=(-1{)}^2,c^2=(-2{)}^2\Rightarrow a^2=9,b^2=1,c^2=4$

Therefore, $a^2+b^2+c^2=9+1+4=14$

Hence, $a^2+b^2+c^2=14$.