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1. I have this question.

I did this,

I divided the whole equation by 5

Then

I know that I'm meant to do something like move the 0.16 to the other side but I'm not too sure.

Help
2. (Original post by zed963)
I have this question.

I did this,

I divided the whole equation by 5

Then

I know that I'm meant to do something like move the 0.16 to the other side but I'm not too sure.

Help
When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e.
3. (Original post by hassi94)
When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e.
I've been watching quite a few videos and guides, I'm getting to the point where I'm confused, I need to stick to one method that works best for me.

The question came from this site.

http://www.mathsisfun.com/algebra/co...ng-square.html

But I'm not fully understanding why they do such things.
4. (Original post by hassi94)
When you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e.
it is not
5. So basically

If I'm correct

I'm guessing that's wrong anyway.
6. How do I carry on this question, I'm basically stuck.
7. (Original post by zed963)
I've been watching quite a few videos and guides, I'm getting to the point where I'm confused, I need to stick to one method that works best for me.

The question came from this site.

http://www.mathsisfun.com/algebra/co...ng-square.html

But I'm not fully understanding why they do such things.
Right well what you're doing when completing the square is adding a little bit which must be taken away again to keep the maths consistent.

say if we want to complete the square on

I've put square brackets around the bit we'll actually be modifying the appearance of.

So we do the regular thing here;

I have written 'adj' to indicate an adjustment that needs to be made, this is because , in fact; as we can clearly see by multiplying out.

We are just trying to get x^2 + 2x and we don't want the '+1' so our adjustment is going to be to take away 1, i.e. adj = -1

So now we have

which we then tidy up to get

since the constant term from the (x+1)^2 is 1, and similarly if it was (x+2)^2 it would be 4 etc. It is always the square of the constant term inside the bracket that is being squared.

So, generally for a quadratic [x^2 + bx] + c, we'd write it as and then the adjustment would be to give us
8. (Original post by raheem94)
it is not
Thanks - just a typo
9. (Original post by zed963)
How do I carry on this question, I'm basically stuck.
Watch this video.
10. (Original post by zed963)
I have this question.

I did this,

I divided the whole equation by 5

Don't know about the method you are using. I'll try to explain the method I find easiest.
If you think of your formula as

Carrying on from the last step I quoted, I use this formula:

x1,2 = - p/2 ± √(p²/4 -q)

Which gives me

x1 = 1.15
x2 = -0.35
11. (Original post by hassi94)
Right well what you're doing when completing the square is adding a little bit which must be taken away again to keep the maths consistent.

say if we want to complete the square on

I've put square brackets around the bit we'll actually be modifying the appearance of.

So we do the regular thing here;

I have written 'adj' to indicate an adjustment that needs to be made, this is because , in fact; as we can clearly see by multiplying out.

We are just trying to get x^2 + 2x and we don't want the '+1' so our adjustment is going to be to take away 1, i.e. adj = -1

So now we have

which we then tidy up to get

since the constant term from the (x+1)^2 is 1, and similarly if it was (x+2)^2 it would be 4 etc. It is always the square of the constant term inside the bracket that is being squared.

So, generally for a quadratic [x^2 + bx] + c, we'd write it as and then the adjustment would be to give us
I understand everything else apart from this.

So, generally for a quadratic [x^2 + bx] + c, we'd write it as and then the adjustment would be to give us

12. (Original post by qua)
Don't know about the method you are using. I'll try to explain the method I find easiest.
If you think of your formula as

Carrying on from the last step I quoted, I use this formula:

x1,2 = - p/2 ± √(p²/4 -q)

Which gives me

x1 = 1.15
x2 = -0.35

When did you learn this?

It seems like this method was from the 90s no offence.
13. (Original post by qua)
And it still works wonders.

And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways.
Oh right no wonder I wasn't familiar with it.
14. (Original post by qua)
And it still works wonders.

And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways.
That is just a less general form of the quadratic equation and does not help the OP complete the square.

(Original post by zed963)
I understand everything else apart from this.

So, generally for a quadratic [x^2 + bx] + c, we'd write it as and then the adjustment would be to give us

What don't you understand about it? I'm saying if we have any quadratic x^2 + bx + c we would complete the square as follows. b and c just represent constant numbers, in the example I showed you above, b = 2 and c = 2.
15. (Original post by hassi94)
That is just a less general form of the quadratic equation and does not help the OP complete the square.

What don't you understand about it? I'm saying if we have any quadratic x^2 + bx + c we would complete the square as follows. b and c just represent constant numbers, in the example I showed you above, b = 2 and c = 2.
I'm just wondering where you have got the (x+b/2)^2 from
16. (Original post by zed963)
I'm just wondering where you have got the (x+b/2)^2 from
like when we have x^2 + 4x, we do (x+2)^2, always putting half of the x coefficient in the bracket. So if we have x^2 + bx, we'd put it as (x+b/2)^2

Make sense?
17. (Original post by hassi94)
like when we have x^2 + 4x, we do (x+2)^2, always putting half of the x coefficient in the bracket. So if we have x^2 + bx, we'd put it as (x+b/2)^2

Make sense?
Oh right I understand I thought it was something much more complicated.

Thanks

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