# Homework Help: Equation problem 2

1. Nov 27, 2009

### 1/2"

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!!

2. Nov 27, 2009

### Staff: Mentor

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.

3. Nov 27, 2009

### 1/2"

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.

4. Nov 27, 2009

### Staff: Mentor

(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.

5. Nov 28, 2009

### 1/2"

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:

6. Nov 28, 2009

### Staff: Mentor

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?

7. Nov 28, 2009

### 1/2"

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

8. Nov 28, 2009

### 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?

9. Nov 28, 2009

### 1/2"

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

10. Nov 28, 2009

### 1/2"

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!

11. Nov 28, 2009

### Staff: Mentor

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...

12. Nov 28, 2009

Thanks!!