Completing the square

I have this question.


$5x^2 – 4x – 2 = 0$

I did this,

I divided the whole equation by 5

$x^2-0.8x-0.4=0$

Then

$(x-0.4x)^2.....=0$

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. $x^2 - 0.8x = (x-0.4x)^2 - 0.16$

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. $x^2 - 0.8x = (x-0.4x)^2 - 0.16$

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/completing-square.html

But I'm not fully understanding why they do such things.

it is $(x-0.4)^2 \ldots$ not $(x-0.4x)^2 \ldots$

How do I carry on this question, I'm basically stuck.

I have this question.


$5x^2 – 4x – 2 = 0$

I did this,

I divided the whole equation by 5

$x^2-0.8x-0.4=0$

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

$[x^2 + 2x] + 2$

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

So we do the regular thing here;

$= [(x+1)^2 + adj] + 2$

I have written 'adj' to indicate an adjustment that needs to be made, this is because $(x+1)^2 \neq x^2 + 2x$, in fact; $(x+1)^2 = x^2 + 2x + 1$ 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

$[(x+1)^2 - 1] + 2$ which we then tidy up to get $(x+1)^2 + 1$

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 $[(x+\frac{b}{2})^2 + adj] + c$ and then the adjustment would be $-\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4}$ to give us $[x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c$

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 $x^2+px+q=0$

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

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.

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.

I understand everything else apart from this.

So, generally for a quadratic [x^2 + bx] + c, we'd write it as $[(x+\frac{b}{2})^2 + adj] + c$ and then the adjustment would be $-\left(\frac{b}{2}\right)^2 = -\frac{b^2}{4}$ to give us $[x^2+ bx] + c = (x+\frac{b}{2})^2 -\frac{b^2}{4} + c$

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

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?