Completing the square
Maths and statistics discussion, revision, exam and homework help.
-
- Reputation:
- TSR Idol
- Location: Yorkshire
Re: Completing the squareWhen you change x^2 - 0.8x to a 'completed square' format - you're adding 0.16 so you must take it away, i.e.(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
Last edited by Intriguing Alias; 01-07-2012 at 16:26. -
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the squareI'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.(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.
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. -
- Reputation:
- TSR Demigod
- Posts: 5,512
Re: Completing the squareit is(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.
not 
-
- Reputation:
- TSR Idol
- Location: Yorkshire
Re: Completing the squareRight 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.(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.
say if we want to complete the square on
![[x^2 + 2x] + 2](http://www.thestudentroom.co.uk/latexrender/pictures/7f/7f86b8e3220e6e57c5d00e72bdd25fe4.png "[x^2 + 2x] + 2")
I've put square brackets around the bit we'll actually be modifying the appearance of.
So we do the regular thing here;
![= [(x+1)^2 + adj] + 2](http://www.thestudentroom.co.uk/latexrender/pictures/f2/f28545c204b413cd9604f92194ac141b.png "= [(x+1)^2 + adj] + 2")
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
![[(x+1)^2 - 1] + 2](http://www.thestudentroom.co.uk/latexrender/pictures/f5/f5e1aeeb34bcc7b0a9dbc5240575eb70.png "[(x+1)^2 - 1] + 2")
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![[(x+\frac{b}{2})^2 + adj] + c](http://www.thestudentroom.co.uk/latexrender/pictures/59/5965249396a2713dbf48f85bc3e4c29d.png "[(x+\frac{b}{2})^2 + adj] + c")
and then the adjustment would be 
to give us ![[x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c](http://www.thestudentroom.co.uk/latexrender/pictures/0e/0ebe0e53b2aef013b39a24ab3c8a5553.png "[x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c")
Last edited by Intriguing Alias; 01-07-2012 at 16:28. -
- Reputation:
- TSR Demigod
- Posts: 5,512
Re: Completing the squareWatch this video.(Original post by zed963)
How do I carry on this question, I'm basically stuck. -
- Reputation:
- Exalted Member
- Posts: 333
Re: Completing the squareDon'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 = -0.35Last edited by qua; 01-07-2012 at 16:32. -
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the squareI understand everything else apart from this.(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
So, generally for a quadratic [x^2 + bx] + c, we'd write it as and then the adjustment would be to give us
-
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the square(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. -
- Reputation:
- Exalted Member
- Posts: 333
Re: Completing the squareAnd it still works wonders.(Original post by zed963)
When did you learn this?
It seems like this method was from the 90s no offence.
And when did I learn it.. 2008? I'm not from the UK though. Just find it the easiest and quickest of ways. -
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the squareOh right no wonder I wasn't familiar with it.(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. -
- Reputation:
- TSR Idol
- Location: Yorkshire
Re: Completing the squareThat is just a less general form of the quadratic equation and does not help the OP complete the square.(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.
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.(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
-
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the squareI'm just wondering where you have got the (x+b/2)^2 from(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. -
- Reputation:
- TSR Idol
- Location: Yorkshire
Re: Completing the squarelike 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(Original post by zed963)
I'm just wondering where you have got the (x+b/2)^2 from
Make sense? -
- Reputation:
- Thread Starter
- Overlord in Training
- Posts: 2,134
Re: Completing the squareOh right I understand I thought it was something much more complicated.(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?
Thanks
