find all values of a, b , and c so that ......

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nfirstyear

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find all values of a, b , and c so that a^2+b^2+c^2+2(2b-a-4c)=-21

note ^2means squared
how do you do that on the forum?
 
The way you wrote square is just fine.
Can you please tell us what progress you made with this problem. It is silly for us to tell you how to start such a problem when you need help at the end of the problem. In you show us your work, as the forum guidelines requires, then we know how to help you. In fact if you followed the forum rules you would have received help by now. Please post back with your work.
 
( a^2 + 2a) + (b^2 - 4b) + (c^2+ 8c) = -21

Calculate the min of each : ( a^2 + 2a), (b^2 - 4b) and (c^2+ 8c)

Minimum (c2 + 8c)= -8 (by derivation of x2+8x we get the min=-8 obtained when x = -4)

I let you continue.
 
I would complete the squares in ( a^2 - 2a) + (b^2 + 4b) + (c^2 - 8c) = -21
The answer should be obvious at that point
 
kongzi said:
( a^2 + 2a) + (b^2 - 4b) + (c^2+ 8c) = -21

Calculate the min of each : ( a^2 + 2a), (b^2 - 4b) and (c^2+ 8c)

Minimum (c2 + 8c)= -8 (by derivation of x2+8x we get the min=-16 obtained when x = -4)

I let you continue.
Click to expand...

Mistake corrected.
 
I strongly recommend Jomo's suggestion of completing the three squares! Have you tried it?
 
To continue with Prof Khan's post.

(x+a)^2 = x^2 + 2ax + a^2. Not that if you take 1/2 of the coefficient of the x-term--that is compute (1/2) of 2a-- and then square that result you get a^2 which is what is needed to be added to get a perfect square.

For example if you have x^2 + 6x. (1/2) of 6 is 3 and 3^2=9. So x^2 + 6x + 9 is a perfect square, name (x+3)^2
 
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