Question

If \[{a^2} + {b^2} + {c^2} + 3 = 2(a + b + c)\] then the value of \[(a + b + c)\] is:
A) 2
B) 3
C) 4
D) 5

Answer 1

Hint: Try to get it in the form of \[{x^2} + {y^2} - 2xy = {(x - y)^2}\] this algebraic formula. and also remember that the sum of squares is always greater than or equal to 0 i.e., it cannot be negative.
Complete step by step answer:
 Let us try to transform the equation a bit
We can rewrite it as
\[\begin{array}{l}
{a^2} + {b^2} + {c^2} + 3 = 2(a + b + c)\\
 \Rightarrow {a^2} + {b^2} + {c^2} - 2(a + b + c) + 3 = 0\\
 \Rightarrow {a^2} + {b^2} + {c^2} - 2a - 2b - 2c + 1 + 1 + 1 = 0\\
 \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + {c^2} - 2c + 1 = 0
\end{array}\]
Now the formula \[{x^2} + {y^2} - 2xy = {(x - y)^2}\] we can write \[x = a,b,c\] simultaneously and \[y = 1\] then we will get the transformed equation as
\[ \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} = 0\]
Now all three of them has to be individually 0 or the combined result cannot be 0 in any case for all real number
Therefore we are getting 3 relations i.e.,
\[\begin{array}{l}
{\left( {a - 1} \right)^2} = 0\\
 \Rightarrow a - 1 = 0\\
 \Rightarrow a = 1\\
{\left( {b - 1} \right)^2} = 0\\
 \Rightarrow b - 1 = 0\\
 \Rightarrow b = 1\\
{\left( {c - 1} \right)^2} = 0\\
 \Rightarrow c - 1 = 0\\
 \Rightarrow c = 1\\
\therefore a = b = c = 1
\end{array}\]
 As we are getting this therefore \[a + b + c = 1 + 1 + 1 = 3\]
Therefore option (B) is correct

Note: It is true for any real number the squared terms cannot be negative because if a negative sign is present in a term after squaring it it will automatically becomes positive but it is not true for complex number because for any complex number \[i = \sqrt { - 1} \] if we will square it we will get \[{i^2} = - 1\] which indeed is a negative number. So it is only true for the real number line.