Equation problem 2

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1. The problem statement, all variables and given/known data
The problem is a 2+b 2+c 2=2(a-b-c)-3
Find a+b+c


2. Relevant equations
none


3. 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!!

[Mark44]

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.

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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.

[Mark44]

(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.

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(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??:uhh:
So what do I do after this??:uhh:

[Mark44]

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?

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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?

[ideasrule]

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?

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Well i cant guess it any more(SORRY!!)
plz work out the next step

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Do you mean to say :eek:
(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":confused:
And ideasrule also makes a point.
I 'm a bit confused !!
I would be happy if you help!

[Borek]

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...

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Thanks!!