# Solve Equation Problem 2: Find a+b+c

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In summary, the problem is a 2+b 2+c 2=2(a-b-c)-3 that can be solved for a, b, and c if the brackets are opened.
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## Homework Statement

The problem is a 2+b 2+c 2=2(a-b-c)-3
Find a+b+c

none

## The Attempt at a Solution

well i tried to add 2(ab+ac+cb) to make the a 2+b 2+c 2=(a+b+c) 2
so the other side also has a 2(ab+ac+cb)
:. it becomes( if brackets are opened)
2a-2b-2c+2ab+2ac+2cb-2-1
=2a+2ab+2ac-2b-2c-2+2cb-1
=2a(1+b+c)-2(1+b+c) +2cb-1
= (1+b+c)(2a-2)+2cb-1
Hey i even close to any proper solution
Plz help me out!

a2 + b2 + c2 = 2a - 2b - 2c - 3
<==> a2 - 2a + b2 + 2b + c2 + 2c = -3

Now complete the square in the a terms, b terms and c terms. You should be able to solve for a, b, and c after that.

Mark44 said:
a2 + b2 + c2 = 2a - 2b - 2c - 3
<==> a2 - 2a + b2 + 2b + c2 + 2c = -3

Now complete the square in the a terms, b terms and c terms. You should be able to solve for a, b, and c after that.

I am sorry but I dont't get it.
Could you please be more elobrate.It would be rather better if you could just work out only 1 more step.

(a2 -2a + ?) + (b2 + 2b + ?) + (c2 + 2c + ?) = -3
Complete the square in each parenthesized group. Whatever you add on the left side, make sure you add the same amount on the right side.

Mark44 said:
(a2 -2a + ?) + (b2 + 2b + ?) + (c2 + 2c + ?) = -3
Complete the square in each parenthesized group. Whatever you add on the left side, make sure you add the same amount on the right side.
I've tried it like this
(a 2-2a+1)+(b2 + 2b + 1) +(c2 + 2c + 1) = -3 +1+1+1
<==>(a-1)2+(b+1)2+(c+1)2=0
Is that ok??or is it wrong??
So what do I do after this??

Yes, that's right, but you're not quite done. The square of any real number is >= 0. So for example, (a - 1)^2 >= 0, and if (a - 1)^2 > 0, what does that say about the other two terms on the left side of the equation?

Mark44 said:
Yes, that's right, but you're not quite done. The square of any real number is >= 0. So for example, (a - 1)^2 >= 0, and if (a - 1)^2 > 0, what does that say about the other two terms on the left side of the equation?

So does that mean
(b+1)>0 and (c+1)>0(??)
But what does that conclude?

Look at the equation (a-1)2+(b+1)2+(c+1)2=0: it says the sum of three squares is 0. None of the squares can be negative. If even one of the squares is positive, what happens?

Well i can't guess it any more(SORRY!)
please work out the next step

Do you mean to say
(a-1)^2=0
=>a-1=0
=>a=1
Similarly b=-1 c=-1
and so a+b+c= -1
But I still have a doubt why can't we consider (a-1)^2= 3or 45 or 1or any positive integer
cuz you said
"The square of any real number is >= 0"
And ideasrule also makes a point.
I 'm a bit confused !
I would be happy if you help!

If (a-1)^2 = 3 you would need one of the other expressions to be negative for the sum to be 0 - but squares can be only positive...

Thanks!

## 1. What is the purpose of solving equation problem 2?

The purpose of solving equation problem 2 is to find the sum of three variables, a, b, and c, by solving the equation a + b + c = 0.

## 2. What is the first step in solving equation problem 2?

The first step in solving equation problem 2 is to gather all the terms with variables on one side of the equation and constants on the other side.

## 3. What should I do if the equation has fractions or decimals?

If the equation has fractions or decimals, you should first eliminate them by multiplying both sides of the equation by the least common multiple of all the denominators.

## 4. How do I know if my solution is correct?

You can check your solution by substituting the values of a, b, and c into the original equation and seeing if it equals to 0. If it does, then your solution is correct.

## 5. Can I solve equation problem 2 using different methods?

Yes, there are multiple methods for solving equation problem 2, such as elimination, substitution, and graphing. It is important to choose the method that you feel most comfortable with and that works best for the given equation.

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