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# How would you complete the square of these two expressions?

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1. Can someone please walk me through and explain the process of completing the square of these expressions:

1) x^2 + 4x -5

2) x^2 - 20x +10

Thanks!
2. (Original post by blobbybill)
Can someone please walk me through and explain the process of completing the square of these expressions:

1) x^2 + 4x -5

2) x^2 - 20x +10

Thanks!
Do you know where to begin? Have you tried anything?
3. (Original post by RDKGames)
Do you know where to begin? Have you tried anything?
I know you halve the co-efficient of X and put it in a bracket, then square the bracket, but I don't know how that would work if one of the signs was negative and another was positive (eg the first question)
4. (Original post by blobbybill)
I know you halve the co-efficient of X and put it in a bracket, then square the bracket, but I don't know how that would work if one of the signs was negative and another was positive (eg the first question)
In exactly the same way. Remember that so subtracting a number is the same as adding the negative of it.

So the first one would be just as you said:

Can you work out and as a result work out the completed square form from there?
5. (Original post by blobbybill)
Can someone please walk me through and explain the process of completing the square of these expressions:

1) x^2 + 4x -5

2) x^2 - 20x +10

Thanks!
Completing the square requires you to do the same formulaic thing each time. In terms of a, b and c, the first thing you do is;

(x+b/2)^2. -- So in your first question the first part you'd have is '(x+2)^2

The next thing you do is square your b/2 term (you always write this number as a negative);

(x+b/2)^2 - b/2^2 -- so in your question it'd be (x+2)^2 - 4

And the last thing you do is add on the c term;

(x+b/2)^2 - b/2^2 + c -- so in your question it'd be (x+2)^2 - 4 - 5 which equals (x+2)^2 - 9

And that's the completed square ^

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