If \[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} = 2\left( {{\text{a + 2b - 2c}}} \right) - 9\]. Then find the value of algebraic expression a+b+c.
Answer
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Hint: Proceed the solution of this question, by breaking 9 into 4 + 4 + 1 then on arranging terms we can apply identities [ ${\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$ & ${\left( {{\text{a + b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}}} \right)$]
Complete step-by-step solution -
In the question it Is given that
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} = 2\left( {{\text{a + 2b - 2c}}} \right) - 9\]
{By opening the brackets and bringing the terms to one side}
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} - 2{\text{a - 4b + 4c}} + 9 = 0\]
Now you can see that we can write 9 = 4 + 4 + 1
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} - 2{\text{a - 4b + 4c}} + 4 + 4 + 1 = 0\] ……. (1)
In order to make whole squares of (a-1)², (b-2)², (c+2)², arrange the terms of equation (1) accordingly
(a² -2a + 1) + (b²– 4b +4) + (c² + 4c +4) =0 ……. (2)
We know the identities [ ${\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$ & ${\left( {{\text{a + b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}}} \right)$]
So applying above identities on equation (2)
⇒ (a-1) ² + (b-2) ²+ (c+2) ²= 0
Now we know (a-1)², (b-2)², (c+2)² are positive terms and greater than or equal to 0 for all real values of a, b, c.
Here the sum of all square terms is so we conclude that it is possible when all the terms individually equals to 0.
Hence we can equalise all to zero.
⇒ (a-1) ² =0 ⇒ a=1
⇒ (b-2) ²=0 ⇒ b=2
⇒ (c+2) ²= 0 ⇒ c=-2
From here we get a = 1; b= 2 and c= -2
On Putting values of a, b, c in
⇒ a+b+c = 1+2-2 = 1
Note: Whenever we come up with such types of questions, so here we need to visualise any certain identity for that we might have to do some short of replacement. Along with that we should know that the sum of squares terms is either positive or zero. So if any equation containing sum of squares terms to zero, then it is only possible when each individual square will have to be 0.
Complete step-by-step solution -
In the question it Is given that
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} = 2\left( {{\text{a + 2b - 2c}}} \right) - 9\]
{By opening the brackets and bringing the terms to one side}
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} - 2{\text{a - 4b + 4c}} + 9 = 0\]
Now you can see that we can write 9 = 4 + 4 + 1
\[{{\text{a}}^2} + {{\text{b}}^2} + {{\text{c}}^2} - 2{\text{a - 4b + 4c}} + 4 + 4 + 1 = 0\] ……. (1)
In order to make whole squares of (a-1)², (b-2)², (c+2)², arrange the terms of equation (1) accordingly
(a² -2a + 1) + (b²– 4b +4) + (c² + 4c +4) =0 ……. (2)
We know the identities [ ${\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$ & ${\left( {{\text{a + b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} + 2{\text{ab}}} \right)$]
So applying above identities on equation (2)
⇒ (a-1) ² + (b-2) ²+ (c+2) ²= 0
Now we know (a-1)², (b-2)², (c+2)² are positive terms and greater than or equal to 0 for all real values of a, b, c.
Here the sum of all square terms is so we conclude that it is possible when all the terms individually equals to 0.
Hence we can equalise all to zero.
⇒ (a-1) ² =0 ⇒ a=1
⇒ (b-2) ²=0 ⇒ b=2
⇒ (c+2) ²= 0 ⇒ c=-2
From here we get a = 1; b= 2 and c= -2
On Putting values of a, b, c in
⇒ a+b+c = 1+2-2 = 1
Note: Whenever we come up with such types of questions, so here we need to visualise any certain identity for that we might have to do some short of replacement. Along with that we should know that the sum of squares terms is either positive or zero. So if any equation containing sum of squares terms to zero, then it is only possible when each individual square will have to be 0.
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