Question & Answer
QUESTION

If ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3=2\left( a-b-c \right)$, then the value of \[2a-b+c\] is:
A. 3
B. 4
C. 0
D. 2

ANSWER Verified Verified
Hint: Rearrange the expression, ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3=2\left( a-b-c \right)$ and then we try if any identity can be applied here. Since, we have three variables a, b and c. So, the constant here 3 is split as (1+1+1).
Then, we will apply the identity, ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\ and\ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Then, we will use the property that the sum of all three real quantities is zero, if all are zero. To get the required values of a, b and c. Then, using calculation, we will find the value of $2a-b+c$.


Complete step-by-step answer:
We have the expression, ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+3=2\left( a-b-c \right)$.
Transferring 3 to the right hand side, we get,
${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2\left( a-b-c \right)-3$
Expanding right hand side, we get,
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2a-2b-2c-3$
Transferring every term from right hand side to left hand side, we get,
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a+2b+2c+3=0$
Since, we have three variables a, b and c. So, we split 3 as (1+1+1), then we get,
$\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a+2b+2c+1+1+1=0..........\left( 1 \right)$
Here, we are trying to arrange the terms in such a way to apply the identity here.
In order to apply the identity,
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\ and\ \\
 & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
\end{align}$
We will arrange the equation (1) as,
$\Rightarrow \left( {{a}^{2}}-2a+1 \right)+\left( {{b}^{2}}+2b+1 \right)+\left( {{c}^{2}}+2c+1 \right)=0$
Now, we can easily transform the terms into simplest one by using the identities,
$\Rightarrow {{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}}+{{\left( c+1 \right)}^{2}}=0$
Using sum of all three real quantity is zero, if all are zero, we get,
$\begin{align}
  & \Rightarrow a-1=0 \\
 & \Rightarrow a=1 \\
 & and \\
 & \Rightarrow b+1=0 \\
 & \Rightarrow b=\left( -1 \right) \\
 & and \\
 & \Rightarrow c+1=0 \\
 & \Rightarrow c=\left( -1 \right) \\
\end{align}$
The values we get for a, b and c are 1, -1 and -1 respectively.
To find the value of $2a-b+c$, we will put the value of a, b and c in the expression.
Hence,
$\begin{align}
  & 2a-b+c=2\left( 1 \right)-\left( -1 \right)+\left( -1 \right) \\
 & =2+1-1 \\
 & =3-1 \\
 & =2 \\
\end{align}$
Hence, option D is correct.

Note: Students might start solving the question by expanding and not focussing on constant term 3, which needs to be split as (1+1+1) to convert the expression into identity.
Here, we need to look 3 as 1+1+1. Since, we have three variables a, b and c.