Question

If $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2\left( a+2b-2c \right)-9 $ then find the value of $ a+b+c $ .
A. 2
B. 3
C. 1
D. None of these

Answer 1

Hint: First you have terms of a, b, c separately. So, try to convert the whole sum into a sum of squares. If the sum of squares is zero then all must be zero. Use this condition to find values of a, b, c. then add them to get their sum. This sum is required to result in this question.

Complete step-by-step answer:
The given equation in the question is written in form of:
 $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2\left( a+2b-2c \right)-9 $
By multiplying the 2 inside bracket, we get:
 $ \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2a+4b-4c-9 $
By subtracting 2a on both sides, we get:
 $ \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a=4b-4c-9 $
By subtracting 4b on both sides, we get:
 $ \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a-4b=-4c-9 $
By adding 4c on both sides, we get
 $ \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a-4b+4c=-9 $
By adding 9 on both sides, we get:
  $ \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a-4b+4c+9=0 $
By combining similar terms together, we get it in form:
 $ \Rightarrow \left( {{a}^{2}}-2a \right)+\left( {{b}^{2}}-4b \right)+\left( {{c}^{2}}+4c \right)+9=0 $
By basic knowledge of algebra we know the formula given by:
 $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}},{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} $
We can write our equation in the form as given below:
 $ \Rightarrow \left( {{a}^{2}}-2\left( a \right)\left( 1 \right) \right)+\left( {{b}^{2}}-2\left( b \right)\left( 2 \right) \right)+\left( {{c}^{2}}+\left( 2 \right)\left( c \right)\left( 2 \right) \right)+9=0 $
By dividing 9 as 1, 4, 4 and given it to each term, we get:
 $ \Rightarrow \left( {{a}^{2}}-2a+1 \right)+\left( {{b}^{2}}-2b+4 \right)+\left( {{c}^{2}}+4c+4 \right)=0 $
By using formula given above, we can write it as:
 $ \Rightarrow {{\left( a-1 \right)}^{2}}+{{\left( b-2 \right)}^{2}}+{{\left( c+2 \right)}^{2}}=0 $
We know the square of a number is always $ \ge 0 $ .
So, for their sum being 0, all of them must be 0.
By equating first term to 0, we get the equation as:
 $ a-1=0 $
By adding 1 on both sides, we get it in form of:
 $ \Rightarrow a=1 $ ………………….. (1)
By equation second term to 0, we get it in the form of:
 $ b-2=0 $
By adding 2 on both sides, we get it in the form of:
 $ \Rightarrow b=2 $ ………………….. (2)
By equating third term to 0, we get it in the form of:
 $ c+2=0 $
By substituting 2 on both sides, we get it in the form of:
 $ \Rightarrow c=-2 $ …………………… (3)
By adding equations (1), (2), (3) we get it in the form of:
 $ \Rightarrow a+b+c=1+2-2=1 $
Therefore, option (c) is the correct option for a given question.

Note: Be careful while getting equation terms on the left hand side. If you miss any sign, then you may not be able to convert it into a sum of squares. If you are not able to convert then you might not get a result. While turning them into square check signs ‘+’ or ‘-‘ you are using.